Exercise 6.1: Simple Simultaneous Equations in Two Variables
(a) Solve the following by using the method of substitution(1-12):
1. Solution: Given equations,
$x + y = 4 \quad \dots \text{(i)}$
$x – y = 2 \quad \dots \text{(ii)}$
From equation (i), we get
$x = 4 – y \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$(4 – y) – y = 2$
or, $4 – 2y = 2$
or, $-2y = 2 – 4$
or, $-2y = -2$
or, $y = 1$
Substituting the value of $y$ into equation (iii), we get:
$x = 4 – 1 = 3$
Required solution: $(x, y) = (3, 1)$
2. Solution: Given equations,$
2x + y = 5 \quad \dots \text{(i)}$
$x – y = 1 \quad \dots \text{(ii)}$
From equation (ii), we get:
$x = y + 1 \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (i), we get:
$2(y + 1) + y = 5$
or, $2y + 2 + y = 5$
or, $3y = 5 – 2$
or, $3y = 3$
or, $y = 1$
Substituting the value of $y$ into equation (iii), we get:
$x = 1 + 1 = 2$
Required solution: $(x, y) = (2, 1)$
3. Solution: Given equations,
$3x + 2y = 10 \quad \dots \text{(i)}$
$x – y = 0 \quad \dots \text{(ii)}$
From equation (ii), we get
$x = y \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (i), we get:
$3y + 2y = 10$
or, $5y = 10$
or, $y = 2$
Substituting the value of $y$ into equation (iii), we get:
$x = 2$
Required solution: $(x, y) = (2, 2)$
4. Solution: Given equations,
$\frac{x}{a} + \frac{y}{b} = \frac{1}{a} + \frac{1}{b} \quad \dots \text{(i)}$
$\frac{x}{a} – \frac{y}{b} = \frac{1}{a} – \frac{1}{b} \quad \dots \text{(ii)}$
From equation (ii), we get
$\frac{x}{a} = \frac{1}{a} – \frac{1}{b} + \frac{y}{b} \quad \dots \text{(iii)}$
Substituting the value of $\frac{x}{a}$ into equation (i), we get:
$\left(\frac{1}{a} – \frac{1}{b} + \frac{y}{b}\right) + \frac{y}{b} = \frac{1}{a} + \frac{1}{b}$
or, $-\frac{1}{b} + \frac{2y}{b} = \frac{1}{b}$
or, $\frac{2y}{b} = \frac{1}{b} + \frac{1}{b}$
or, $\frac{2y}{b} = \frac{2}{b}$
or, $y = 1$
Substituting the value of $y$ into equation (iii), we get:
$\frac{x}{a} = \frac{1}{a} – \frac{1}{b} + \frac{1}{b}$
or, $\frac{x}{a} = \frac{1}{a}$
or, $x = 1$
Required solution: $(x, y) = (1, 1)$
5. Solution: Given equations,
$3x – 2y = 0 \quad \dots \text{(i)}$
$17x – 7y = 13 \quad \dots \text{(ii)}$
From equation (i), we get
$3x = 2y$
or, $x = \frac{2y}{3} \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$17\left(\frac{2y}{3}\right) – 7y = 13$
or, $\frac{34y}{3} – 7y = 13$
or, $\frac{34y – 21y}{3} = 13$
or, $\frac{13y}{3} = 13$
or, $13y = 39$
or, $y = 3$
Substituting the value of $y$ into equation (iii), we get:
$x = \frac{2 \times 3}{3} = 2$
Required solution: $(x, y) = (2, 3)$
6. Solution: Given equations,
$x – y = 2a \quad \dots \text{(i)}$
$ax + by = a^2 + b^2 \quad \dots \text{(ii)}$
From equation (i), we get
$x = 2a + y \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$a(2a + y) + by = a^2 + b^2$
or, $2a^2 + ay + by = a^2 + b^2$
or, $y(a + b) = a^2 + b^2 – 2a^2$
or, $y(a + b) = b^2 – a^2$
or, $y(a + b) = (b – a)(b + a)$
or, $y = b – a$
Substituting the value of $y$ into equation (iii), we get:
$x = 2a + (b – a) = a + b$
Required solution: $(x, y) = (a + b, b – a)$
7. Solution: Given equations,
$ax + by = ab \quad \dots \text{(i)}$
$bx + ay = ab \quad \dots \text{(ii)}$
From equation (i), we get
$ax = ab – by$
or, $x = \frac{ab – by}{a} \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$b\left(\frac{ab – by}{a}\right) + ay = ab$
or, $\frac{ab^2 – b^2 y}{a} + ay = ab$
or, $\frac{ab^2 – b^2 y + a^2 y}{a} = ab$
or, $ab^2 + y(a^2 – b^2) = a^2 b$
or, $y(a^2 – b^2) = a^2 b – ab^2$
or, $y(a – b)(a + b) = ab(a – b)$
or, $y = \frac{ab}{a + b}$
Substituting the value of $y$ into equation (iii), we get:
$x = \frac{ab – b\left(\frac{ab}{a + b}\right)}{a}$
or, $x = \frac{\frac{ab(a + b) – ab^2}{a + b}}{a}$
or, $x = \frac{\frac{a^2 b + ab^2 – ab^2}{a + b}}{a}$
or, $x = \frac{\frac{a^2 b}{a + b}}{a} = \frac{ab}{a + b}$
Required solution: $(x, y) = \left(\frac{ab}{a + b}, \frac{ab}{a + b}\right)$
8. Solution: Given equations,
$ax – by = ab \quad \dots \text{(i)}$
$bx – ay = ab \quad \dots \text{(ii)}$
From equation (i), we get,
$ax = ab + by$
or, $x = \frac{ab + by}{a} \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$b\left(\frac{ab + by}{a}\right) – ay = ab$
or, $\frac{ab^2 + b^2 y}{a} – ay = ab$
or, $\frac{ab^2 + b^2 y – a^2 y}{a} = ab$
or, $ab^2 – y(a^2 – b^2) = a^2 b$
or, $-y(a^2 – b^2) = a^2 b – ab^2$
or, $-y(a^2 – b^2) = ab(a – b)$
or, $y(a – b)(a + b) = -ab(a – b)$
or, $y = -\frac{ab}{a + b}$
Substituting the value of $y$ into equation (iii), we get:
$x = \frac{ab + b\left(-\frac{ab}{a + b}\right)}{a}$
or, $x = \frac{\frac{ab(a + b) – ab^2}{a + b}}{a}$
or, $x = \frac{\frac{a^2 b + ab^2 – ab^2}{a + b}}{a}$
or, $x = \frac{ab}{a + b}$
Required solution: $(x, y) = \left(\frac{ab}{a + b}, -\frac{ab}{a + b}\right)$
9. Solution: Given equations,
$ax – by = a – b \quad \dots \text{(i)}$
$ax + by = a + b \quad \dots \text{(ii)}$
From equation (i), we get
$ax = a – b + by$
or, $x = \frac{a – b + by}{a} \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get:
$a\left(\frac{a – b + by}{a}\right) + by = a + b$
or, $a – b + by + by = a + b$
or, $a – b + 2by = a + b$
or, $2by = a + b – a + b$
or, $2by = 2b$
or, $y = 1$
Substituting the value of $y$ into equation (iii), we get:
$x = \frac{a – b + b(1)}{a} = \frac{a}{a} = 1$
Required solution: $(x, y) = (1, 1)$
Required solution: $(x, y) = (1, 1)$
10. Solution: Given equations,
$\frac{1}{x} + \frac{1}{y} = \frac{5}{6} \quad \dots \text{(i)}$
$\frac{1}{x} – \frac{1}{y} = \frac{1}{6} \quad \dots \text{(ii)}$
Let, $\frac{1}{x} = u$ and $\frac{1}{y} = v$
Then the equations become,
$u + v = \frac{5}{6} \quad \dots \text{(iii)}$
$u – v = \frac{1}{6} \quad \dots \text{(iv)}$
From equation (iv), we get
$u = \frac{1}{6} + v \quad \dots \text{(v)}$
Substituting the value of $u$ into equation (iii), we get:
$\left(\frac{1}{6} + v\right) + v = \frac{5}{6}$
or, $2v = \frac{5}{6} – \frac{1}{6}$
or, $2v = \frac{4}{6} = \frac{2}{3}$
or, $v = \frac{1}{3}$
Substituting the value of $v$ into equation (v), we get
$u = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}$
Now,
$\frac{1}{x} = u = \frac{1}{2} \Rightarrow x = 2$
$\frac{1}{y} = v = \frac{1}{3} \Rightarrow y = 3$
Required solution: $(x, y) = (2, 3)$
11. Solution: Given equations,
$\frac{x}{a} + \frac{y}{b} = \frac{2}{a} + \frac{1}{b} \quad \dots \text{(i)}$
$\frac{x}{b} – \frac{y}{a} = \frac{2}{b} – \frac{1}{a} \quad \dots \text{(ii)}$
From equation (i), we get
$\frac{x}{a} = \frac{2}{a} + \frac{1}{b} – \frac{y}{b}$
or, $x = a\left(\frac{2}{a} + \frac{1}{b} – \frac{y}{b}\right) = 2 + \frac{a}{b} – \frac{ay}{b} \quad \dots \text{(iii)}$
Substituting this value of $x$ into equation (ii), we get
$\frac{1}{b}\left(2 + \frac{a}{b} – \frac{ay}{b}\right) – \frac{y}{a} = \frac{2}{b} – \frac{1}{a}$
or, $\frac{2}{b} + \frac{a}{b^2} – \frac{ay}{b^2} – \frac{y}{a} = \frac{2}{b} – \frac{1}{a}$
or, $\frac{a}{b^2} + \frac{1}{a} = \frac{ay}{b^2} + \frac{y}{a}$
or, $\frac{a^2 + b^2}{ab^2} = y\left(\frac{a^2 + b^2}{ab^2}\right)$
or, $y = 1$
Substituting the value of $y$ into equation (iii), we get:
$x = 2 + \frac{a}{b} – \frac{a(1)}{b} = 2$
Required solution: $(x, y) = (2, 1)$
12. Solution: Given equations,
$\frac{a}{x} + \frac{b}{y} = \frac{a}{2} + \frac{b}{3} \quad \dots \text{(i)}$
$\frac{a}{x} – \frac{b}{y} = \frac{a}{2} – \frac{b}{3} \quad \dots \text{(ii)}$
From equation (ii), we get
$\frac{a}{x} = \frac{a}{2} – \frac{b}{3} + \frac{b}{y} \quad \dots \text{(iii)}$
Substituting the value of $\frac{a}{x}$ into equation (i), we get
$\left(\frac{a}{2} – \frac{b}{3} + \frac{b}{y}\right) + \frac{b}{y} = \frac{a}{2} + \frac{b}{3}$
or, $-\frac{b}{3} + \frac{2b}{y} = \frac{b}{3}$
or, $\frac{2b}{y} = \frac{b}{3} + \frac{b}{3}$
or, $\frac{2b}{y} = \frac{2b}{3}$
or, $y = 3$
Substituting the value of $y$ into equation (iii), we get:
$\frac{a}{x} = \frac{a}{2} – \frac{b}{3} + \frac{b}{3}$
or, $\frac{a}{x} = \frac{a}{2}$
or, $x = 2$
Required solution: $(x, y) = (2, 3)$
(b) Solve the following by using the method of elimination(13-26):
13. Solution: Given equations,
$x – y = 4 \quad \dots \text{(i)}$
$x + y = 6 \quad \dots \text{(ii)}$
Adding equations (i) and (ii), we get
$(x – y) + (x + y) = 4 + 6$
or, $2x = 10$
or, $x = 5$
Substituting the value of $x$ into equation (ii), we get
$5 + y = 6$
or, $y = 6 – 5$
or, $y = 1$
Required solution: $(x, y) = (5, 1)$
14. Solution: Given equations,
$2x + 3y = 7 \quad \dots \text{(i)}$
$6x – 7y = 5 \quad \dots \text{(ii)}$
Multiplying equation (i) by 3, we get
$6x + 9y = 21 \quad \dots \text{(iii)}$
Subtracting equation (ii) from equation (iii), we get
$(6x + 9y) – (6x – 7y) = 21 – 5$
or, $16y = 16$
or, $y = 1$
Substituting the value of $y$ into equation (i), we get:
$2x + 3(1) = 7$
or, $2x + 3 = 7$
or, $2x = 4$
or, $x = 2$
Required solution: $(x, y) = (2, 1)$
15. Solution: Given equations,
$4x + 3y = 15 \quad \dots \text{(i)}$
$5x + 4y = 19 \quad \dots \text{(ii)}$
Multiplying equation (i) by 4 and (ii) by 3, we get
$16x + 12y = 60 \quad \dots \text{(iii)}$
$15x + 12y = 57 \quad \dots \text{(iv)}$
Subtracting equation (iv) from (iii), we get
$(16x + 12y) – (15x + 12y) = 60 – 57$
or, $x = 3$
Substituting the value of $x$ into equation (i), we get
$4(3) + 3y = 15$
or, $12 + 3y = 15$
or, $3y = 3$
or, $y = 1$
Required solution: $(x, y) = (3, 1)$
16. Solution: Given equations,
$3x – 2y = 5 \quad \dots \text{(i)}$
$2x + 3y = 12 \quad \dots \text{(ii)}$
Multiplying equation (i) by 3 and (ii) by 2, we get
$9x – 6y = 15 \quad \dots \text{(iii)}$
$4x + 6y = 24 \quad \dots \text{(iv)}$
Adding equations (iii) and (iv), we get:
$13x = 39$
or, $x = 3$
Substituting the value of $x$ into equation (i), we get
$3(3) – 2y = 5$
or, $9 – 2y = 5$
or, $-2y = 5 – 9$
or, $-2y = -4$
or, $y = 2$
Required solution: $(x, y) = (3, 2)$
17. Solution: Given equations,
$4x – 3y = -1 \quad \dots \text{(i)}$
$3x – 2y = 0 \quad \dots \text{(ii)}$
Multiplying equation (i) by 2 and (ii) by 3, we get
$8x – 6y = -2 \quad \dots \text{(iii)}$
$9x – 6y = 0 \quad \dots \text{(iv)}$
Subtracting equation (iii) from (iv), we get
$(9x – 6y) – (8x – 6y) = 0 – (-2)$
or, $x = 2$
Substituting the value of $x$ into equation (ii), we get
$3(2) – 2y = 0$
or, $6 – 2y = 0$
or, $2y = 6$
or, $y = 3$
Required solution: $(x, y) = (2, 3)$
18. Solution: Given equations,
$3x – 5y = -9 \quad \dots \text{(i)}$
$5x – 3y = 1 \quad \dots \text{(ii)}$
Multiplying equation (i) by 3 and (ii) by 5, we get
$9x – 15y = -27 \quad \dots \text{(iii)}$
$25x – 15y = 5 \quad \dots \text{(iv)}$
Subtracting equation (iii) from (iv), we get
$(25x – 15y) – (9x – 15y) = 5 – (-27)$
or, $16x = 32$
or, $x = 2$
Substituting the value of $x$ into equation (ii), we get
$5(2) – 3y = 1$
or, $10 – 3y = 1$
or, $-3y = -9$
or, $y = 3$
Required solution: $(x, y) = (2, 3)$
19. Solution: Given equations,
$\frac{x}{2} + \frac{y}{2} = 3 \quad \dots \text{(i)}$
$\frac{x}{2} – \frac{y}{2} = 1 \quad \dots \text{(ii)}$
Adding equations (i) and (ii), we get
$\left(\frac{x}{2} + \frac{y}{2}\right) + \left(\frac{x}{2} – \frac{y}{2}\right) = 3 + 1$
or, $\frac{2x}{2} = 4$
or, $x = 4$
Subtracting equation (ii) from (i), we get
$\left(\frac{x}{2} + \frac{y}{2}\right) – \left(\frac{x}{2} – \frac{y}{2}\right) = 3 – 1$
or, $\frac{2y}{2} = 2$
or, $y = 2$
Required solution: $(x, y) = (4, 2)$
20. Solution: Given equations,
$x + ay = b \quad \dots \text{(i)}$
$ax – by = c \quad \dots \text{(ii)}$
Multiplying equation (i) by $a$, we get
$ax + a^2 y = ab \quad \dots \text{(iii)}$
Subtracting equation (ii) from (iii), we get
$(ax + a^2 y) – (ax – by) = ab – c$
or, $a^2 y + by = ab – c$
or, $y(a^2 + b) = ab – c$
or, $y = \frac{ab – c}{a^2 + b}$
Multiplying equation (i) by $b$ and equation (ii) by $a$, we get
$bx + aby = b^2 \quad \dots \text{(iv)}$
$a^2 x – aby = ac \quad \dots \text{(v)}$
Adding equations (iv) and (v), we get
$bx + a^2 x = b^2 + ac$
or, $x(a^2 + b) = ac + b^2$
or, $x = \frac{ac + b^2}{a^2 + b}$
Required solution: $(x, y) = \left(\frac{ac + b^2}{a^2 + b}, \frac{ab – c}{a^2 + b}\right)$
21. Solution: Given equations,
$\frac{x}{2} + \frac{y}{3} = 3 \quad \dots \text{(i)}$
$x – \frac{y}{3} = 3 \quad \dots \text{(ii)}$
Adding equations (i) and (ii), we get
$\left(\frac{x}{2} + \frac{y}{3}\right) + \left(x – \frac{y}{3}\right) = 3 + 3$
or, $\frac{x}{2} + x = 6$
or, $\frac{3x}{2} = 6$
or, $3x = 12$
or, $x = 4$
Substituting the value of $x$ into equation (ii), we get
$4 – \frac{y}{3} = 3$
or, $-\frac{y}{3} = 3 – 4$
or, $-\frac{y}{3} = -1$
or, $y = 3$
Required solution: $(x, y) = (4, 3)$
22. Solution: Given equations,
$\frac{x}{3} + \frac{2}{y} = 1 \quad \dots \text{(i)}$
$\frac{x}{4} – \frac{3}{y} = 3 \quad \dots \text{(ii)}$
Multiplying equation (i) by 3 and (ii) by 2, we get
$x + \frac{6}{y} = 3 \quad \dots \text{(iii)}$
$\frac{x}{2} – \frac{6}{y} = 6 \quad \dots \text{(iv)}$
Adding equations (iii) and (iv), we get
$x + \frac{x}{2} = 3 + 6$
or, $\frac{3x}{2} = 9$
or, $3x = 18$
or, $x = 6$
Substituting the value of $x$ into equation (i), we get
$\frac{6}{3} + \frac{2}{y} = 1$
or, $2 + \frac{2}{y} = 1$
or, $\frac{2}{y} = 1 – 2$
or, $\frac{2}{y} = -1$
or, $y = -2$
Required solution: $(x, y) = (6, -2)$
23. Solution: Given equations,
$\frac{x}{a} + \frac{y}{b} = \frac{2}{a} + \frac{1}{b} \quad \dots \text{(i)}$
$\frac{x}{b} – \frac{y}{a} = \frac{2}{b} – \frac{1}{a} \quad \dots \text{(ii)}$
Multiplying equation (i) by $\frac{1}{a}$ and (ii) by $\frac{1}{b}$, we get
$\frac{x}{a^2} + \frac{y}{ab} = \frac{2}{a^2} + \frac{1}{ab} \quad \dots \text{(iii)}$
$\frac{x}{b^2} – \frac{y}{ab} = \frac{2}{b^2} – \frac{1}{ab} \quad \dots \text{(iv)}$
Adding equations (iii) and (iv), we get
$\frac{x}{a^2} + \frac{x}{b^2} = \frac{2}{a^2} + \frac{2}{b^2}$
or, $x\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 2\left(\frac{1}{a^2} + \frac{1}{b^2}\right)$
or, $x = 2$
Multiplying equation (i) by $\frac{1}{b}$ and (ii) by $\frac{1}{a}$, we get
$\frac{x}{ab} + \frac{y}{b^2} = \frac{2}{ab} + \frac{1}{b^2} \quad \dots \text{(v)}$
$\frac{x}{ab} – \frac{y}{a^2} = \frac{2}{ab} – \frac{1}{a^2} \quad \dots \text{(vi)}$
Subtracting equation (vi) from (v), we get
$\frac{y}{b^2} + \frac{y}{a^2} = \frac{1}{b^2} + \frac{1}{a^2}$
or, $y\left(\frac{1}{a^2} + \frac{1}{b^2}\right) = 1 \cdot \left(\frac{1}{a^2} + \frac{1}{b^2}\right)$
or, $y = 1$
Required solution: $(x, y) = (2, 1)$
24. Solution: Given equations,
$\frac{a}{x} + \frac{b}{y} = \frac{a}{2} + \frac{b}{3} \quad \dots \text{(i)}$
$\frac{a}{x} – \frac{b}{y} = \frac{a}{2} – \frac{b}{3} \quad \dots \text{(ii)}$
Adding equations (i) and (ii), we get
$\left(\frac{a}{x} + \frac{b}{y}\right) + \left(\frac{a}{x} – \frac{b}{y}\right) = \left(\frac{a}{2} + \frac{b}{3}\right) + \left(\frac{a}{2} – \frac{b}{3}\right)$
or, $\frac{2a}{x} = \frac{2a}{2}$
or, $\frac{2a}{x} = a$
or, $ax = 2a$
or, $x = 2$
Subtracting equation (ii) from (i), we get
$\left(\frac{a}{x} + \frac{b}{y}\right) – \left(\frac{a}{x} – \frac{b}{y}\right) = \left(\frac{a}{2} + \frac{b}{3}\right) – \left(\frac{a}{2} – \frac{b}{3}\right)$
or, $\frac{2b}{y} = \frac{2b}{3}$
or, $y = 3$
Required solution: $(x, y) = (2, 3)$
25. Solution: Given equations,
$\frac{x}{6} + \frac{2}{y} = 2 \quad \dots \text{(i)}$
$\frac{x}{4} – \frac{1}{y} = 1 \quad \dots \text{(ii)}$
Multiplying equation (ii) by 2, we get
$\frac{x}{2} – \frac{2}{y} = 2 \quad \dots \text{(iii)}$
Adding equations (i) and (iii), we get
$\frac{x}{6} + \frac{x}{2} = 2 + 2$
or, $\frac{x + 3x}{6} = 4$
or, $\frac{4x}{6} = 4$
or, $4x = 24$
or, $x = 6$
Substituting the value of $x$ into equation (i), we get
$\frac{6}{6} + \frac{2}{y} = 2$
or, $1 + \frac{2}{y} = 2$
or, $\frac{2}{y} = 2 – 1$
or, $\frac{2}{y} = 1$
or, $y = 2$
Required solution: $(x, y) = (6, 2)$
26. Solution: Given equations,
$x + y = a – b \quad \dots \text{(i)}$
$ax – by = a^2 + b^2 \quad \dots \text{(ii)}$
Multiplying equation (i) by $b$, we get
$bx + by = ab – b^2 \quad \dots \text{(iii)}$
Adding equations (ii) and (iii), we get
$(ax – by) + (bx + by) = (a^2 + b^2) + (ab – b^2)$
or, $ax + bx = a^2 + ab$
or, $x(a + b) = a(a + b)$
or, $x = a$
Substituting the value of $x$ into equation (i), we get
$a + y = a – b$
or, $y = a – b – a$
or, $y = -b$
Required solution: $(x, y) = (a, -b)$