Key Concepts at a Glance
- A linear equation in two variables has the form ax + by + c = 0 (the standard form), with a, b, c real numbers and a, b not both zero.
- A solution is an ordered pair (x, y) that satisfies the equation. There are infinitely many, and together they form a straight line — its graph.
- Slope = Rise ÷ Run = \(\frac{y_2 - y_1}{x_2 - x_1}\), the same for any two points on the line. It is positive (up to the right), negative (down to the right), 0 (horizontal) or undefined (vertical).
- Slope-intercept form: y = mx + d, where m is the slope and the line crosses the y-axis at (0, d).
- A pair of linear equations can be solved by substitution, elimination or graphically (the point where the two lines meet).
- For \(a_1x + b_1y + c_1 = 0\) and \(a_2x + b_2y + c_2 = 0\): unique solution if \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) (intersecting lines); no solution if \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) (parallel); infinitely many if all three ratios are equal (coincident).
The chapter starts from a simple purchase — mangoes at ₹60 per kg and bananas at ₹50 per kg costing ₹280 in all — which gives the linear equation 60x + 50y = 280. It shows that such an equation has infinitely many solutions, all lying on one straight line, and measures how steep that line is using its slope.
It then puts two equations together: a pair of linear equations usually pins down a single solution, found by substitution, elimination or by drawing both lines. But two lines can also be parallel (no solution) or the same line (infinitely many), and the ratios of the coefficients tell you in advance which case you are in.
In-Text Questions & Think and Reflect
Answers to all seventeen Think and Reflect boxes and the questions the chapter asks inside its explanations, in the order they appear in the textbook.
Think and Reflect
TRAre \(\left(1, \frac{9}{2}\right)\), (0, 6), (4, 1) and \(\left(\frac{1}{2}, \frac{21}{4}\right)\) solutions of the equation 3x + 2y = 12?
Can you find any other solution? How many solutions can you find?
Substitute each pair into 3x + 2y:
\(\left(1, \frac92\right)\): \(3(1) + 2\left(\frac92\right) = 3 + 9 = 12\) — solution.
(0, 6): \(3(0) + 2(6) = 12\) — solution.
(4, 1): \(3(4) + 2(1) = 14 \ne 12\) — not a solution.
\(\left(\frac12, \frac{21}{4}\right)\): \(\frac32 + \frac{21}{2} = \frac{24}{2} = 12\) — solution.
Other solutions: (2, 3), (4, 0), (−2, 9), (6, −3), … Choose any value of x, and \(y = \frac{12 - 3x}{2}\) gives a matching y. So there are infinitely many solutions.
Think and Reflect
TRFour friends—Ranju, Meena, Farhan, and Toshi—are solving problems.
Ranju: I noticed something! If c = 0 in the standard form of a line ax + by + c = 0, then the line must pass through the origin. Look, if I substitute x = 0 and y = 0, in equation ax + by = 0, the equation is satisfied. So, the origin lies on the line! We can also say that the line passes through the origin.
Farhan: Let us try for the equation 2x + 3y = 0. Here a = 2, b = 3 but c = 0. If we substitute x = 0, then we get 3y = 0 or y = 0. This means (0, 0) lies on the line. So yes, this line passes through the origin.
Toshi: Suppose our equation has b = c = 0, say, 5x = 0. This becomes x = 0 which is the equation of the y-axis. And the y-axis passes through the origin.
Meena: And if we take a = c = 0, say, 7y = 0, that means y = 0, which is the equation of the x-axis that also passes through the origin.
So, they conclude: Whenever c = 0, the line ax + by + c = 0 will always pass through the origin, irrespective of the values of a or b. Do you agree with them?
Yes, I agree — with one condition that is already part of the definition: a and b must not both be zero.
If c = 0, the equation is ax + by = 0. Putting x = 0, y = 0 gives a(0) + b(0) = 0, which is true for every a and b. So (0, 0) always lies on the line.
The special cases the friends found fit in: b = c = 0 gives ax = 0, i.e. x = 0 (the y-axis); a = c = 0 gives by = 0, i.e. y = 0 (the x-axis). Both pass through the origin.
If a = b = c = 0, the "equation" is 0 = 0, which is not a line at all — that is why the definition requires a and b not both zero.
Think and Reflect
TR1. Consider the point on the line AB whose x-coordinate is 8. What is the y-coordinate of this point? Express this as an ordered pair.
2. What is the x-coordinate of the point on the line AB whose y-coordinate is 6? Express this as an ordered pair.
Line AB passes through A(2, 0) and B(4, 3), so its slope is \(\frac{3}{2}\). For every 2 units to the right, it rises 3 units.
1. From A(2, 0) to x = 8 is a run of 6 = 3 × 2, so the rise is 3 × 3 = 9. The point is (8, 9) (point D in Fig. 13.8).
2. From A(2, 0) to y = 6 is a rise of 6 = 2 × 3, so the run is 2 × 2 = 4. The x-coordinate is 2 + 4 = 6. The point is (6, 6) (point C).
Check with the equation of the line, \(y = \frac32(x - 2)\): x = 8 gives y = 9, and y = 6 gives x = 6.
Think and Reflect
TRCan you use the points A, B and C to verify that the slope of this line is indeed −5?
Points A, B, C on y = −5x + 3: every pair gives slope −5.
A(0, 3), B(1, −2), C(−1, 8).
Using A and B: \(\dfrac{-2 - 3}{1 - 0} = \dfrac{-5}{1} = -5\)
Using A and C: \(\dfrac{8 - 3}{-1 - 0} = \dfrac{5}{-1} = -5\)
Using B and C: \(\dfrac{8 - (-2)}{-1 - 1} = \dfrac{10}{-2} = -5\)
Think and Reflect
TRThe cost of one paratha and 2 bowls of dahi is ₹90. Also, the cost of 2 parathas and 7 bowls of dahi is ₹260. Can you figure out the cost of one paratha and one bowl of dahi and explain your reasoning?
Let one paratha cost ₹p and one bowl of dahi cost ₹d.
p + 2d = 90 … (1)
2p + 7d = 260 … (2)
Reasoning: doubling order (1) gives 2 parathas and 4 bowls for ₹180. Order (2) has the same 2 parathas but 3 more bowls, and costs ₹260 − ₹180 = ₹80 more. So 3 bowls of dahi cost ₹80.
\(d = \frac{80}{3} \approx 26.67\), and from (1), \(p = 90 - \frac{160}{3} = \frac{110}{3} \approx 36.67\).
Check: \(2 \times \frac{110}{3} + 7 \times \frac{80}{3} = \frac{220 + 560}{3} = 260\). ✓
ITConsider another situation. A science exhibition charges an entry fee that is different for adults and children. Here are the amount paid by two different groups.
Group A: 2 adult tickets and 3 child tickets for ₹600
Group B: 3 adult tickets and 2 child tickets for ₹700
How much do individual adult and child tickets cost?
If the cost of one adult ticket is ₹x and one child ticket is ₹y, then:
2x + 3y = 600, 3x + 2y = 700.
This gives us another pair of linear equations in two variables.
A solution to these equations will give us the answer that we seek. Can you see why?
Why: the actual prices must make both groups' bills correct at the same time. A pair (x, y) that satisfies only one equation fits one group's bill but not the other's. So the true prices are exactly the ordered pair that satisfies both equations.
Solving: adding the equations gives 5x + 5y = 1300, so x + y = 260. Subtracting the first from the second gives x − y = 100.
So x = 180 and y = 80.
Check: 2(180) + 3(80) = 600 ✓, 3(180) + 2(80) = 700 ✓.
Think and Reflect
TRSolve the puzzle and discuss your strategy with your friends. Can you determine the value represented by each shape?
Let T, R, H, C be the values of the triangle, rectangle, hexagon and circle. From the grid:
Row 1: 2T + 2R = 28, so T + R = 14.
Column 2 (R, R, T, R): 3R + T = 30.
Subtracting: 2R = 16, so R = 8 and T = 6.
Row 3 (C, T, C, C): 3C + T = 18, so 3C = 12 and C = 4.
Row 2 (H, R, H, R): 2H + 2R = 30, so H + 8 = 15 and H = 7.
Checks: Row 4: 3C + R = 12 + 8 = 20 ✓. Column 3 (R, H, C, C): 8 + 7 + 4 + 4 = 23 ✓. Column 4 (T, R, C, C): 6 + 8 + 4 + 4 = 22 ✓.
The missing total for Column 1 (T, H, C, C) is 6 + 7 + 4 + 4 = 21.
Think and Reflect
TRWhat if we had expressed y in terms of x, i.e., take Equation (2) and write \(y = \frac{1}{2}(3 - x)\). Would we still get the same solution?
Yes. Substitute \(y = \frac{3 - x}{2}\) into Equation (1), 7x − 15y = 2:
\(7x - \frac{15(3 - x)}{2} = 2\)
\(14x - 45 + 15x = 4\)
\(29x = 49\), so \(x = \frac{49}{29}\).
Then \(y = \frac12\left(3 - \frac{49}{29}\right) = \frac12 \cdot \frac{38}{29} = \frac{19}{29}\).
The pair has only one solution, so whichever variable we express first, we must reach that same solution.
Think and Reflect
TRThe method used in solving this problem is called the Elimination Method because we eliminate one of the variables to obtain a linear equation in the other variable. In the example above, we eliminated y. Try the same problem by eliminating x instead of y. Also try to solve the two equations using the Substitution Method. Assess the pros and cons of each method.
The equations are 9x − 4y = 2000 … (1) and 7x − 3y = 2000 … (2).
Eliminating x: multiply (1) by 7 and (2) by 9:
63x − 28y = 14000 … (3)
63x − 27y = 18000 … (4)
(4) − (3): y = 4000. Then from (1): 9x = 2000 + 16000, so x = 2000.
Substitution: from (2), \(y = \frac{7x - 2000}{3}\). Put this in (1):
\(9x - \frac{4(7x - 2000)}{3} = 2000\) ⇒ 27x − 28x + 8000 = 6000 ⇒ x = 2000, and \(y = \frac{14000 - 2000}{3} = 4000\).
Incomes: 9x = ₹18,000 and 7x = ₹14,000, as before.
Pros and cons: Substitution is natural when one variable already has coefficient 1 (like x + 2y = 3), but here it brings in fractions. Elimination avoids fractions and is quicker when coefficients are easy to match, but may need larger multipliers (63 here). Both always give the same answer.
Think and Reflect
TRDoes a pair of linear equations always have a unique solution? Can you analyse this using the Elimination Method? While subtracting the equations to eliminate a particular variable, if the other variable remains, then clearly there is a unique solution. But what if the second variable gets eliminated too while eliminating the first variable? Can you think of a situation where this happens?
No. Both variables disappear together when, after making the x-coefficients equal, the y-coefficients also become equal — that is, when one equation's left side is a multiple of the other's: \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\).
Example: x − y = 10 and 10x − 10y = 100. Multiplying the first by 10 makes the two equations identical, and subtracting gives 0 = 0. Every solution of x − y = 10 works — infinitely many solutions.
Example: x − y = 10 and 10x − 10y = 101. Now subtracting gives 0 = 1, which is impossible — no solution.
Think and Reflect
TRHow many solutions are there to the equations 10x − 5y = 20 and 4x − 2y = 8?
Give 3 more examples of pairs of equations that lead to 0 = 0 after subtraction, and therefore have infinitely many solutions. Can you find a simple rule to check when this happens?
10x − 5y = 20 is 5(2x − y) = 20, i.e. 2x − y = 4. And 4x − 2y = 8 is 2(2x − y) = 8, i.e. 2x − y = 4. Both are the same equation, so there are infinitely many solutions — every point on 2x − y = 4, e.g. (2, 0), (0, −4), (3, 2).
More examples:
(a) x + y = 2 and 2x + 2y = 4
(b) x − 3y = 1 and 3x − 9y = 3
(c) 2x + 5y = 7 and 6x + 15y = 21
Rule: one equation is a non-zero multiple of the other, i.e. \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\).
Think and Reflect
TRWhat if one of \(a_2, b_2, c_2\) is zero?
Then a ratio like \(\frac{c_1}{c_2}\) cannot be written (division by zero), so we go back to the form the rule came from: the equations have infinitely many solutions exactly when
\(a_1 = k a_2,\ b_1 = k b_2,\ c_1 = k c_2\) for some k ≠ 0.
So a zero coefficient in one equation must be matched by a zero in the same place in the other.
Example: 2x + 3y = 0 and 4x + 6y = 0 (c₁ = c₂ = 0, k = ½) have infinitely many solutions. But 2x + 3y = 5 and 4x + 6y = 0 have c₂ = 0 while c₁ ≠ 0, so they are not multiples; they have no solution.
Similarly, x = 3 and 2x = 6 (b₁ = b₂ = 0) are the same line.
Think and Reflect
TRDoes a pair of linear equations always have either a unique solution or infinitely many solutions?
No — a third case is possible: no solution.
Example: x − y = 10 and 10x − 10y = 101. Multiplying the first by 10 gives 10x − 10y = 100. The left sides are now equal but the right sides are not (100 ≠ 101), so no (x, y) can satisfy both. Geometrically, the two lines are parallel.
Think and Reflect
TRGive 3 more examples of pairs of equations that have no solution. Can you find a simple rule to check when this will happen?
(a) x + y = 1 and x + y = 2
(b) 2x − 3y = 4 and 4x − 6y = 5
(c) x − 5y = 3 and 3x − 15y = 7
In each pair, making the x-coefficients equal also makes the y-coefficients equal, but the constants stay different, so subtracting gives a false statement like 0 = 3.
Rule: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) ⇒ no solution.
Think and Reflect
TRAre the converses of the above statements true?
Yes, all three converses are true.
The three conditions — (1) \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\); (2) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\); (3) \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\) — cover every possible pair, and no pair satisfies two of them. They lead to three different outcomes: unique, infinitely many, none.
So, for example, if a pair has a unique solution, it cannot be in case (2) or (3) (those give infinitely many or none), so it must be in case (1): \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). The same argument works for the other two.
TRCan there be methods other than elimination and substitution to reduce a pair of linear equations in two variables to a linear equation in one variable? If so, describe them.
Yes. A few examples:
Comparison (equating) method: write both equations as y = … and set the two expressions equal. E.g. y = 6 − 2x and y = 2x − 2 give 6 − 2x = 2x − 2, so x = 2.
Adding and subtracting when coefficients are swapped: for 9x + 7y = 107, 7x + 9y = 101, adding gives x + y = 13 and subtracting gives x − y = 3 (Exercise 13.5 Q7).
Cross-multiplication formula: for \(a_1x + b_1y + c_1 = 0\), \(a_2x + b_2y + c_2 = 0\),
(valid when \(a_1b_2 \ne a_2b_1\)). This is elimination done once, in general.
Graphical method: draw both lines and read off the point of intersection (Section 13.6).
ITConsider the following pairs of linear equations in two variables.
Pair 1: 2x + y = 6 and x − y = 2
Pair 2: x + y = 4 and 2x + 2y = 8
Pair 3: 3x − 2y = 6 and 6x − 4y = 12
Pair 4: x − 2y = 4 and 2x − 4y = 6
For each pair, prepare a table of values (with at least two ordered pairs). Plot the points on the Cartesian plane. Draw the straight lines.
Pair 1: 2x + y = 6 through (0, 6), (3, 0); x − y = 2 through (0, −2), (2, 0). The lines intersect at \(\left(\frac83, \frac23\right)\) — adding the equations gives 3x = 8.
Pair 2: x + y = 4 through (0, 4), (4, 0); 2x + 2y = 8 through the same points. The lines are coincident — infinitely many solutions.
Pair 3: 3x − 2y = 6 through (0, −3), (2, 0); 6x − 4y = 12 through the same points. The lines are coincident.
Pair 4: x − 2y = 4 through (0, −2), (4, 0); 2x − 4y = 6 through (0, −1.5), (3, 0). Both have slope \(\frac12\) but different y-intercepts, so the lines are parallel — no solution.
Think and Reflect
TRHow do we find solutions from the graph? What can we say about the number of solutions when the two lines (i) intersect at a point, (ii) are parallel, (iii) are coincident?
A solution of the pair must satisfy both equations, so it is a point lying on both lines. We read off the common point(s) of the two graphs.
(i) Intersecting lines: exactly one common point — a unique solution.
(ii) Parallel lines: no common point — no solution.
(iii) Coincident lines: every point is common — infinitely many solutions.
Think and Reflect
TRCan you prove that two lines of equal slope are parallel? (Hint: Consider the equations of the two lines to be \(y = mx + d_1\) and \(y = mx + d_2\))
Take two different lines with the same slope m: \(y = mx + d_1\) and \(y = mx + d_2\), with \(d_1 \ne d_2\).
Suppose they had a common point (x, y). Then \(mx + d_1 = y = mx + d_2\), so \(d_1 = d_2\) — a contradiction.
So the two lines have no common point, i.e. they are parallel. (If \(d_1 = d_2\), they are the same line.)
Vertical lines (undefined slope) such as x = 2 and x = 5 are also parallel, since no point can have x = 2 and x = 5 at once.
Exercise Set 13.1
Linear equations in two variables and the standard form.
1Write a linear equation in two variables in which a = 3, b = 0 and \(c = -\frac{1}{5}\).
Substituting in ax + by + c = 0:
\(3x + 0 \cdot y - \frac15 = 0\), i.e. \(3x - \frac15 = 0\).
2Complete the following table after expressing the given linear equations in standard form.
| Linear equation | Standard form | Coefficient of x | Coefficient of y | Constant term |
|---|---|---|---|---|
| y − 15 = √2x | √2x − y + 15 = 0 | √2 | −1 | 15 |
| 3y − 2x = 0 | −2x + 3y + 0 = 0 | −2 | 3 | 0 |
| 5x = 3y | 5x − 3y + 0 = 0 | 5 | −3 | 0 |
| x = 8 | x + 0·y − 8 = 0 | 1 | 0 | −8 |
| 3y = 1 | 0·x + 3y − 1 = 0 | 0 | 3 | −1 |
Multiplying a standard form by −1 is also correct (e.g. −√2x + y − 15 = 0 with a = −√2, b = 1, c = −15); the signs of all three values then change together.
3(i) The cost of a notebook is twice the cost of a pen. Consider the cost of a notebook to be ₹t and that of a pen to be ₹p. Charlie wrote t = 2p, whereas Meera wrote p = 2t. Which of these two representations is correct?
(ii) In a one-day International Cricket match between India and Sri Lanka played in Nagpur, two Indian batsmen together scored 176 runs. Manisha expressed this situation as x + y = 176, where the number of runs scored by one batsman is x, and the number of runs scored by the other is y. Is this a correct representation?
(i) Charlie is correct. "Notebook = twice the pen" means t = 2 × p, i.e. t = 2p. Meera's p = 2t would mean the pen costs twice the notebook. (Check: if a pen costs ₹10, a notebook costs ₹20, and 20 = 2 × 10.)
(ii) Yes. The runs of the two batsmen add up to 176, so x + y = 176 (with x and y whole numbers). It is a linear equation in two variables, x + y − 176 = 0.
Exercise Set 13.2
Solutions of a linear equation and its graph.
1Verify if the ordered pair (4, 3) is a solution of 5x − 6y = 2. Explain your reasoning.
Put x = 4, y = 3: 5(4) − 6(3) = 20 − 18 = 2, which equals the right-hand side.
2Find any two solutions for each of the following equations:
(i) 7x − 3y = 21
(ii) 2x + 3y = 5
(i) x = 0 gives −3y = 21, y = −7: (0, −7). y = 0 gives 7x = 21, x = 3: (3, 0).
(ii) x = 1 gives 3y = 3, y = 1: (1, 1). x = −2 gives 3y = 9, y = 3: (−2, 3).
3In the equations given below, m and n are unknown constants: 2mx + 3y = 7; 4x + ny = −10. If (2, −1) is the solution of both equations, find the values of m and n.
First equation: 2m(2) + 3(−1) = 7 ⇒ 4m − 3 = 7 ⇒ 4m = 10 ⇒ \(m = \frac52\).
Second equation: 4(2) + n(−1) = −10 ⇒ 8 − n = −10 ⇒ n = 18.
4Find two solutions which lie in different quadrants for each of the following linear equations. Identify the quadrants in which the points lie.
(i) 5x + 3y = 7
(ii) 5x − 3y = 7
(iii) −5x + 3y = 7
(iv) −5x − 3y = 7
Verify your solutions by representing the linear equations on a graph paper.
(i) (2, −1): 10 − 3 = 7 ✓ — Quadrant IV. (−1, 4): −5 + 12 = 7 ✓ — Quadrant II.
(ii) (2, 1): 10 − 3 = 7 ✓ — Quadrant I. (−1, −4): −5 + 12 = 7 ✓ — Quadrant III.
(iii) (1, 4): −5 + 12 = 7 ✓ — Quadrant I. (−2, −1): 10 − 3 = 7 ✓ — Quadrant III.
(iv) (1, −4): −5 + 12 = 7 ✓ — Quadrant IV. (−2, 1): 10 − 3 = 7 ✓ — Quadrant II.
Plotting each pair of points and joining them, the line passes through both points, confirming the solutions.
5Consider the graph of the equation 3x − 7y = 21 shown below. Does the point C (2, 3) lie on the line? Does it satisfy the equation? Can points that do not lie on the line satisfy the equation?
From the graph, C(2, 3) lies above the line through A(7, 0) and B(0, −3), not on it.
Check: 3(2) − 7(3) = 6 − 21 = −15 ≠ 21, so C does not satisfy the equation.
No point off the line can satisfy the equation: the line is exactly the set of all points whose coordinates satisfy 3x − 7y = 21.
6State whether the following sentences are True or False. Justify your answer.
(i) A linear equation in two variables has only one solution.
(ii) The graph of a linear equation in two variables always passes through the origin.
(iii) A linear equation in two variables can never have rational solutions.
(iv) x = 3 is a valid linear equation in two variables.
(v) The equation 2x + 3y = 7 has infinitely many solutions.
(vi) The point (1, 2) is a solution of the equation 2x + 3y = 7.
(i) False. It has infinitely many solutions; e.g. x + y = 2 has (0, 2), (1, 1), (2, 0), …
(ii) False. Only when c = 0. E.g. x + y = 2: (0, 0) gives 0 ≠ 2.
(iii) False. 3x + 2y = 12 has the rational solutions (0, 6) and \(\left(1, \frac92\right)\).
(iv) True. It can be written as 1·x + 0·y − 3 = 0 with a = 1, b = 0 (a and b not both zero).
(v) True. For every x, \(y = \frac{7 - 2x}{3}\) gives a solution.
(vi) False. 2(1) + 3(2) = 8 ≠ 7.
7(i) Compare the solutions of the equations 3x + 4y = 7 and 6x + 8y = 14. Argue that they have the same set of solutions, that is, every solution of one is also a solution of the other.
(ii) Show that the equations ax + by = c and kax + kby = kc, with k ≠ 0, have the same set of solutions.
(i) If (u, v) satisfies 3u + 4v = 7, then 6u + 8v = 2(3u + 4v) = 2 × 7 = 14, so it satisfies the second equation.
Conversely, if 6u + 8v = 14, dividing by 2 gives 3u + 4v = 7.
So every solution of one is a solution of the other — the same set of solutions (the same line).
(ii) If au + bv = c, then kau + kbv = k(au + bv) = kc.
Conversely, if kau + kbv = kc, then since k ≠ 0 we may divide by k to get au + bv = c.
(If k = 0, the second equation would be 0 = 0, satisfied by every pair — that is why k ≠ 0 is needed.)
Exercise Set 13.3
Slope of a line.
1The following diagrams represent ski hills. Rank the hills in order of their steepness, from least to greatest.
Slope = rise ÷ run (from the diagrams):
Hill A: \(\frac{60}{70} = \frac67 \approx 0.857\)
Hill B: \(\frac{60}{110} = \frac{6}{11} \approx 0.545\)
Hill C: \(\frac{80}{100} = 0.8\)
2The ramp at a loading dock rises 2.5 metres over a run of 4 metres. Find the slope of the ramp.
\(\text{Slope} = \dfrac{\text{Rise}}{\text{Run}} = \dfrac{2.5}{4} = \dfrac{5}{8} = 0.625\)
3Find the slope of the line l in each of the following diagrams.
(i) l is horizontal (parallel to the x-axis), so y does not change: slope = 0.
(ii) l passes through the origin and the marked point (−2, 2) (one grid square = 1 unit): slope \(= \frac{2 - 0}{-2 - 0} = \mathbf{-1}\).
(iii) The marked points are (−2, −1) and (1, 1): slope \(= \frac{1 - (-1)}{1 - (-2)} = \mathbf{\frac23}\).
(iv) l is vertical (parallel to the y-axis): the run is 0, so the slope is undefined.
4An accessibility ramp for wheelchairs is to be made alongside the staircase. The guidelines given for the slope of the ramp is 1 cm vertical rise to 12 cm horizontal length. What should be the horizontal length of the ramp if the total height of the stairs is 18 cm?
Slope \(= \frac{1}{12} = \frac{\text{Rise}}{\text{Run}} = \frac{18}{\text{Run}}\).
Run = 18 × 12 = 216 cm.
Exercise Set 13.4
Given below are some everyday situations. Construct a pair of linear equations in two variables for each situation:
1On two different days a family buys movie tickets and snack boxes. Let the cost of one movie ticket be ₹x and the cost of one snack box be ₹y. Frame a pair of linear equations in x and y to represent the following situations:
(i) On the first day, the family buys 2 movie tickets and 3 snack boxes for ₹850.
(ii) On the second day, the family buys 4 movie tickets and 1 snack box for ₹1100.
(i) 2x + 3y = 850
(ii) 4x + y = 1100
(For checking later: solving gives x = 245, y = 120.)
2Two friends, Sahil and Meena, travelled by taxi. Let the fixed charge for one taxi trip be ₹x and the additional charge per kilometre be ₹y. Frame a pair of linear equations in x and y to represent the following situation:
(i) Sahil travelled 6 km and paid a total fare of ₹122.
(ii) Meena travelled 8 km and paid a total fare of ₹160.
(i) x + 6y = 122
(ii) x + 8y = 160
(For checking later: y = 19, x = 8.)
3In a sports meet, tickets for adults cost ₹150 each and tickets for children cost ₹100 each. A total of 200 people attended the meet, and the total amount collected from ticket sales was ₹25,000. (Let the number of adult tickets sold be x and the number of children's tickets sold be y.) Frame a pair of linear equations in x and y to represent this situation.
Number of people: x + y = 200
Money collected: 150x + 100y = 25000 (or, dividing by 50, 3x + 2y = 500)
(For checking later: x = 100, y = 100.)
Exercise Set 13.5
Solving pairs of linear equations algebraically and graphically.
1Form a pair of linear equations for each of the following problems and find their solutions.
(i) The sum of two integers is +5 and their difference is −21. Find the two numbers.
(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3 bats and 5 balls for ₹4000. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹155 and for a journey of 15 km, the total amount paid is ₹220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes equal to \(\frac{9}{11}\) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to \(\frac{5}{6}\). Find the fraction.
(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to \(\frac{1}{2}\) if we add 1 only to the denominator. What is the fraction?
(vii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(ix) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 did she receive?
(x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
(i) x + y = 5, x − y = −21. Adding: 2x = −16, x = −8; y = 13. The integers are −8 and 13.
(ii) x − y = 26, x = 3y. Then 3y − y = 26, y = 13, x = 39. The numbers are 39 and 13.
(iii) Bat ₹x, ball ₹y: 7x + 6y = 8880, 3x + 5y = 4000. Multiply the first by 5 and the second by 6: 35x + 30y = 44400, 18x + 30y = 24000. Subtracting: 17x = 20400, x = 1200. Then 5y = 4000 − 3600 = 400, y = 80. Bat ₹1200, ball ₹80.
(iv) Fixed charge ₹x, ₹y per km: x + 10y = 155, x + 15y = 220. Subtracting: 5y = 65, y = 13; x = 25. Fixed charge ₹25, ₹13 per km. For 25 km: 25 + 25 × 13 = ₹350.
(v) Fraction \(\frac{x}{y}\): \(\frac{x+2}{y+2} = \frac{9}{11}\) ⇒ 11x − 9y = −4; \(\frac{x+3}{y+3} = \frac56\) ⇒ 6x − 5y = −3. Multiply by 5 and 9: 55x − 45y = −20, 54x − 45y = −27. Subtracting: x = 7; then 9y = 81, y = 9. The fraction is \(\frac79\).
(vi) \(\frac{x+1}{y-1} = 1\) ⇒ x − y = −2; \(\frac{x}{y+1} = \frac12\) ⇒ 2x − y = 1. Subtracting: x = 3, y = 5. The fraction is \(\frac35\).
(vii) Nuri x, Sonu y (present ages): x − 5 = 3(y − 5) ⇒ x − 3y = −10; x + 10 = 2(y + 10) ⇒ x − 2y = 10. Subtracting: y = 20, x = 50. Nuri is 50 years, Sonu is 20 years.
(viii) Tens digit x, units digit y: x + y = 9; 9(10x + y) = 2(10y + x) ⇒ 88x = 11y ⇒ y = 8x. Then 9x = 9, x = 1, y = 8. The number is 18 (9 × 18 = 162 = 2 × 81).
(ix) ₹50 notes x, ₹100 notes y: x + y = 25, 50x + 100y = 2000 ⇒ x + 2y = 40. Subtracting: y = 15, x = 10. 10 notes of ₹50 and 15 notes of ₹100.
(x) Fixed charge ₹x (first three days), ₹y per extra day: x + 4y = 27 (7 days), x + 2y = 21 (5 days). Subtracting: 2y = 6, y = 3; x = 15. Fixed charge ₹15, ₹3 per extra day.
2Form a pair of linear equations and find their common solutions graphically.
10 students of Grade 9 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
x + y = 10 and y = x + 4 meet at (3, 7): 3 boys and 7 girls.
Let the number of boys be x and girls be y.
x + y = 10: points (0, 10), (10, 0), (5, 5).
y = x + 4: points (0, 4), (2, 6), (6, 10).
The lines meet at (3, 7). Check: 3 + 7 = 10 ✓ and 7 = 3 + 4 ✓.
3Use the ratios \(\frac{a_1}{a_2}, \frac{b_1}{b_2}, \frac{c_1}{c_2}\), to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident.
(i) 5x − 4y + 8 = 0; 7x + 6y − 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0; 2x − y + 9 = 0
(i) \(\frac{a_1}{a_2} = \frac57\), \(\frac{b_1}{b_2} = \frac{-4}{6} = -\frac23\). These are unequal, so the lines intersect at a point.
(ii) \(\frac{9}{18} = \frac12\), \(\frac{3}{6} = \frac12\), \(\frac{12}{24} = \frac12\). All equal, so the lines are coincident.
(iii) \(\frac62 = 3\), \(\frac{-3}{-1} = 3\), \(\frac{10}{9} \ne 3\). So the lines are parallel.
4Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x − y = 8, 3x − 3y = 16
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
(i) \(\frac12 = \frac12 = \frac{5}{10}\): coincident lines — infinitely many solutions, every point of x + y = 5, e.g. (0, 5), (2, 3), (5, 0).
(ii) \(\frac13 = \frac{-1}{-3} \ne \frac{8}{16}\): parallel lines — no solution.
(iii) \(\frac24 \ne \frac{1}{-2}\): the lines intersect — a unique solution. The second equation is 2x − y = 2.
2x + y = 6 and 2x − y = 2 intersect at (2, 2).
2x + y = 6 through (0, 6), (3, 0); 2x − y = 2 through (0, −2), (1, 0). They meet at (2, 2).
(iv) \(\frac24 = \frac{-2}{-4} = \frac12\) but \(\frac{-2}{-5} = \frac25 \ne \frac12\): parallel lines — no solution.
5Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Length l, width w: l + w = 36 (half the perimeter), l = w + 4.
Then (w + 4) + w = 36 ⇒ 2w = 32 ⇒ w = 16, l = 20.
6Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables so that the graphs of the pair formed represent (i) intersecting lines (ii) parallel lines (iii) coincident lines.
(i) Intersecting: need \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\). E.g. x − y − 1 = 0 (\(\frac21 \ne \frac{3}{-1}\)).
(ii) Parallel: need \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\). E.g. 4x + 6y − 7 = 0 (\(\frac24 = \frac36 = \frac12\), but \(\frac{-8}{-7} \ne \frac12\)).
(iii) Coincident: multiply by any non-zero number. E.g. 4x + 6y − 16 = 0.
7Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (c. 850 CE). The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of each citron and of each fragrant wood-apple. (Hint: The given problem can be modelled as
9x + 7y = 107
7x + 9y = 101
Can you figure out a way of solving these equations without directly using elimination or the substitution method? What is special in this pair of equations? The x and y coefficients are interchanged in the 2 equations.
What will happen if we add the pair of equations? What will happen if we subtract the pair of equations? Can the resulting equations be solved to find the values of x and y?)
9x + 7y = 107 … (1) and 7x + 9y = 101 … (2)
Add: 16x + 16y = 208 ⇒ x + y = 13 … (3)
Subtract (2) from (1): 2x − 2y = 6 ⇒ x − y = 3 … (4)
Adding (3) and (4): 2x = 16, x = 8. Then y = 5.
Check: 9(8) + 7(5) = 72 + 35 = 107 ✓; 7(8) + 9(5) = 56 + 45 = 101 ✓.
Because the coefficients are interchanged, adding and subtracting give two very simple equations for x + y and x − y.
85 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.
Pencil ₹x, pen ₹y: 5x + 7y = 50, 7x + 5y = 46.
Adding: 12x + 12y = 96 ⇒ x + y = 8. Subtracting: 2y − 2x = 4 ⇒ y − x = 2.
So y = 5, x = 3. Check: 15 + 35 = 50 ✓, 21 + 25 = 46 ✓.
9Find the height of the stool.
Let the stool's height be s cm and the cat's height be c cm.
Cat sitting on the stool: s + c = 85.
Cat on the floor beside the stool, 25 cm below the top of the stool: s − c = 25.
Adding: 2s = 110, s = 55 (and c = 30).
End-of-Chapter Exercises
All 16 End-of-Chapter Exercises. Starred (*) questions are the more challenging ones.
1The graph of the line y = 3x, passing through (0, 0) and (2, 6) is given below.
(i) Identify the slope of the line from the graph and explain how you calculated it using the two points on the line.
(ii) What is the y-intercept of the line? Explain its significance.
(iii) A water tank is being filled so that the water level y (in cm) after x minutes follows this graph.
(a) How high is the water after 5 minutes?
(b) How many minutes will it take for the water to reach a height of 21 cm?
(iv) Without drawing a new graph, determine whether the point (4, 10) lies on this line. Justify your answer.
(v) Plot the point where this line intersects the line x = 3. Explain how you found the coordinates.
(i) From (0, 0) to (2, 6): rise = 6, run = 2, so slope \(= \frac{6 - 0}{2 - 0} = 3\).
(ii) The y-intercept is 0 — the line passes through the origin. For the tank, it means the water level is 0 cm at the start (x = 0).
(iii)(a) y = 3 × 5 = 15 cm.
(b) 21 = 3x ⇒ x = 7 minutes.
(iv) For x = 4 the line gives y = 12, not 10. So (4, 10) does not lie on the line.
(v) On x = 3, every point has x-coordinate 3. Substituting in y = 3x gives y = 9. The lines meet at (3, 9).
2In countries like the USA, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Celsius to Fahrenheit: \(F = \frac{9}{5}C + 32\)
(i) Draw the graph of the linear equation above using Celsius on the x-axis and Fahrenheit on the y-axis.
(ii) If the temperature is 30 °C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95 °F, what is the temperature in Celsius?
(iv) If the temperature is 0 °C, what is the temperature in Fahrenheit and if the temperature is 0 °F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
(i) Table: C = 0 → F = 32; C = 100 → F = 212; C = −40 → F = −40.
Celsius on the x-axis, Fahrenheit on the y-axis. The line passes through (0, 32), (100, 212) and (−40, −40).
(ii) \(F = \frac95 \times 30 + 32 = 54 + 32 = 86\). So 30 °C = 86 °F.
(iii) \(95 = \frac95C + 32\) ⇒ \(\frac95C = 63\) ⇒ C = 35. So 95 °F = 35 °C.
(iv) C = 0 gives F = 32 °F. F = 0 gives \(\frac95C = -32\), so \(C = -\frac{160}{9} \approx\) −17.8 °C.
(v) Put F = C: \(C = \frac95C + 32\) ⇒ \(-\frac45C = 32\) ⇒ C = −40. So −40 °C = −40 °F. On the graph, this is where the line meets y = x.
3Solve the following system of equations graphically: 2x + y = 6, 2x − y − 2 = 0.
2x + y = 6 and 2x − y = 2 intersect at (2, 2).
2x + y = 6: points (0, 6), (3, 0).
2x − y = 2: points (0, −2), (1, 0).
The lines intersect at (2, 2). Check: 4 + 2 = 6 ✓, 4 − 2 − 2 = 0 ✓.
4Find the point of intersection of the lines shown on the cover page.
The cover shows two lines: the x-axis (the horizontal white line with the arrow marked x) and the slanted line labelled 15x − 10y − 115 = 0.
Every point on the x-axis has y = 0. Substituting y = 0 in the equation of the slanted line:
15x − 10(0) − 115 = 0 ⇒ 15x = 115 ⇒ \(x = \frac{115}{15} = \frac{23}{3} \approx 7.67\)
So the lines intersect at \(\left(\frac{23}{3}, 0\right)\).
Check using the slope-intercept form: 10y = 15x − 115 gives \(y = \frac32x - \frac{23}{2}\). This line has slope \(\frac32\) and y-intercept \(-\frac{23}{2}\), and at \(x = \frac{23}{3}\), \(y = \frac32 \cdot \frac{23}{3} - \frac{23}{2} = 0\). ✓
(The second, nearly horizontal orange light beam on the cover has no equation printed on it, so its intersection with the labelled line cannot be calculated from the cover.)
5Give a formula to find the x-intercept of the line y = mx + c.
The line meets the x-axis where y = 0: 0 = mx + c ⇒ \(x = -\frac{c}{m}\) (for m ≠ 0).
So the line cuts the x-axis at \(\left(-\frac{c}{m}, 0\right)\).
If m = 0, the line y = c is horizontal: it has no x-intercept when c ≠ 0, and it is the x-axis itself when c = 0.
6A person is choosing between two mobile plans.
Plan A: ₹50 monthly fee + ₹0.20 per minute of call time.
Plan B: ₹30 monthly fee + ₹0.30 per minute of call time.
For how many minutes of calling per month is Plan A cheaper than Plan B? For how many minutes is Plan B cheaper? Also find the number of minutes at which both plans cost the same.
For t minutes: Plan A costs y = 50 + 0.2t; Plan B costs y = 30 + 0.3t.
Equal cost: 50 + 0.2t = 30 + 0.3t ⇒ 20 = 0.1t ⇒ t = 200. Both cost 50 + 40 = ₹90.
For more than 200 minutes, Plan B's higher per-minute rate makes it costlier, so Plan A is cheaper. For fewer than 200 minutes, Plan B is cheaper (e.g. 100 minutes: A = ₹70, B = ₹60).
7How many lines exist that
(i) have a given slope?
(ii) have a given slope and pass through a given point?
(i) Infinitely many. y = mx + d for every value of d — a family of parallel lines.
(ii) Exactly one. If the line y = mx + d passes through \((x_1, y_1)\), then \(y_1 = mx_1 + d\) fixes \(d = y_1 - mx_1\). (Two such lines would be parallel with a common point, so they would coincide.)
8For what values of p does the pair of equations given below have a unique solution? 4x + py + 8 = 0; 2x + 2y + 2 = 0.
Unique solution ⇔ \(\frac{a_1}{a_2} \ne \frac{b_1}{b_2}\) ⇔ \(\frac42 \ne \frac{p}{2}\) ⇔ p ≠ 4.
For p = 4: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = 2\) but \(\frac{c_1}{c_2} = \frac82 = 4\), so the lines are parallel and there is no solution.
9Find the values of a and b for which the following system of equations has infinitely many solutions: (a + b)x − 2by = 5a + 2b + 1; 3x − y = 14.
We need \(\frac{a+b}{3} = \frac{-2b}{-1} = \frac{5a + 2b + 1}{14}\).
From \(\frac{a+b}{3} = 2b\): a + b = 6b ⇒ a = 5b.
From \(2b = \frac{5a + 2b + 1}{14}\): 28b = 5a + 2b + 1 = 25b + 2b + 1 ⇒ b = 1.
So b = 1, a = 5. Check: the first equation becomes 6x − 2y = 28, which is 2(3x − y) = 2 × 14 ✓.
10Find the value of ‘k’ for which the following system of equations represents a pair of coincident lines: x + 2y = 3; (k − 1)x + (k + 1)y = k + 3.
We need \(\frac{1}{k-1} = \frac{2}{k+1} = \frac{3}{k+3}\).
From the first two: k + 1 = 2(k − 1) ⇒ k = 3.
Check with k = 3: \(\frac12 = \frac24 = \frac36\) ✓. The second equation is 2x + 4y = 6 = 2(x + 2y = 3).
11(i) Robot 1 starts from the origin and traces a path by repeatedly moving 3 units to the right and then 4 units upward. Robot 2 starts from the point (10, 0) and repeatedly moves 1 unit to the right and then 2 units upward. Will the paths traced by these two robots intersect and if so, where?
(ii) Robot 1 starts from (3, 0) and traces a path by repeatedly moving 5 units to the right and then 5 units upward. Robot 2 starts from the point (7, 0) and repeatedly moves 5 units to the right and then 3 units downwards. Will the paths traced by these two robots intersect and if so, where?
(i) After k complete moves, Robot 1 is at (3k, 4k); these points lie on \(y = \frac43x\). After j complete moves, Robot 2 is at (10 + j, 2j); these lie on y = 2(x − 10), i.e. y = 2x − 20.
Solving: \(\frac43x = 2x - 20\) ⇒ \(\frac23x = 20\) ⇒ x = 30, y = 40.
Robot 1 reaches (30, 40) after k = 10 moves, and Robot 2 reaches it after j = 20 moves. Yes — the paths meet at (30, 40).
(If you draw the actual zig-zag steps, they also touch along a few short stretches near this point, but (30, 40) is the point both robots reach at the end of a complete move, where the two lines cross.)
(ii) Robot 1's positions (3 + 5k, 5k) lie on y = x − 3. Robot 2's positions (7 + 5j, −3j) lie on \(y = -\frac35(x - 7)\).
These lines meet where \(x - 3 = -\frac35(x - 7)\) ⇒ 5x − 15 = −3x + 21 ⇒ x = 4.5, y = 1.5. But Robot 2 never reaches this point: it starts at x = 7 and only moves right, with y ≤ 0.
After its first move, Robot 1 is always above the x-axis (y ≥ 5), while Robot 2 is on or below it. The only overlap is at the very start: Robot 1's first move goes along the x-axis from (3, 0) to (8, 0), passing over Robot 2's starting point (7, 0). So the paths share only the short stretch from (7, 0) to (8, 0) on the x-axis; beyond that they never meet.
12*At a certain time, Jacob notices that his digital watch reads ‘a’ minutes after two o’clock. Fifteen minutes later, it reads ‘b’ minutes after three o’clock. He noticed that a is six times greater than b. What time was it when he looked at his watch for the second time?
From 2:a to 3:00 takes (60 − a) minutes, then b more minutes to 3:b. So (60 − a) + b = 15 ⇒ a − b = 45.
Also a = 6b. So 6b − b = 45 ⇒ b = 9, a = 54.
First look: 2:54. Second look (15 minutes later): 3:09.
13*The sum of the digits of a two-digit number is 15. The number obtained by interchanging the digits exceeds the given number by 9. Find the number.
Tens digit x, units digit y: x + y = 15.
(10y + x) − (10x + y) = 9 ⇒ 9(y − x) = 9 ⇒ y − x = 1.
So y = 8, x = 7. The number is 78 (87 − 78 = 9 ✓).
14*In a cyclic quadrilateral ABCD, ∠A = (x + 7)°, ∠B = (y + 8)°, ∠C = (3y + 23)° and ∠D = (4x + 12)°. Find all four angles of the cyclic quadrilateral.
Opposite angles of a cyclic quadrilateral add up to 180°.
∠A + ∠C = 180: x + 7 + 3y + 23 = 180 ⇒ x + 3y = 150 … (1)
∠B + ∠D = 180: y + 8 + 4x + 12 = 180 ⇒ 4x + y = 160 … (2)
From (2), y = 160 − 4x. In (1): x + 480 − 12x = 150 ⇒ 11x = 330 ⇒ x = 30, y = 40.
∠A = 37°, ∠B = 48°, ∠C = 143°, ∠D = 132°. (37 + 143 = 180 ✓, 48 + 132 = 180 ✓)
15*A train moving with uniform speed for a certain distance takes 6 hours less if its speed is increased by 6 km/hour. It would have taken 6 hours more had its speed been decreased by 4 km/hour. Find the distance travelled and the speed of the train.
Let the speed be v km/h and the time t hours; distance = vt.
(v + 6)(t − 6) = vt ⇒ −6v + 6t − 36 = 0 ⇒ t − v = 6 … (1)
(v − 4)(t + 6) = vt ⇒ 6v − 4t − 24 = 0 ⇒ 3v − 2t = 12 … (2)
From (1), t = v + 6. In (2): 3v − 2v − 12 = 12 ⇒ v = 24, t = 30.
Distance = 24 × 30 = 720 km. Check: 30 × 24 = 720 ✓ and 20 × 36 = 720 ✓.
16*The age of a father is equal to the sum of the ages of his four children. After 20 years, the sum of the ages of the children will be twice the age of the father. Find the age of the father.
Let the father's age be F and the children's total age be S. Then F = S.
After 20 years, each of the 4 children is 20 years older, so the total becomes S + 80, and the father is F + 20.
S + 80 = 2(F + 20) ⇒ F + 80 = 2F + 40 ⇒ F = 40.
Extra Practice Questions
Seven extra questions for independent practice once you have gone through the solved questions above. Try each one, then tap to check your answer.
1Find two solutions of 3x − 4y = 12.
x = 0 gives y = −3; y = 0 gives x = 4.
2Find the slope of the line through (2, 5) and (6, 13).
\(\dfrac{13 - 5}{6 - 2} = \dfrac{8}{4} = 2\)
3Find k if (2, −1) is a solution of 3x + ky = 8.
3(2) + k(−1) = 8 ⇒ 6 − k = 8 ⇒ k = −2.
4Write 4x + 2y = 10 in slope-intercept form and state its slope and y-intercept.
2y = −4x + 10 ⇒ y = −2x + 5.
5Solve by elimination: 3x + 2y = 11, 5x − 2y = 13.
Adding: 8x = 24, x = 3. Then 2y = 11 − 9 = 2, y = 1.
6Without solving, decide whether 2x − 3y = 5 and 6x − 9y = 20 have a unique solution, no solution or infinitely many.
\(\frac26 = \frac{-3}{-9} = \frac13\), but \(\frac{5}{20} = \frac14 \ne \frac13\).
7The sum of two numbers is 50 and one is 10 more than the other. Find the numbers.
x + y = 50, x − y = 10 ⇒ 2x = 60, x = 30, y = 20.