Exam practice

From the second, y = 2x − 12. Then
$$\begin{aligned} 3x + 5(2x - 12) &= 31 \\ 13x - 60 &= 31 \\ x &= 7 \end{aligned}$$
y = 14 − 12 = 2. x = 7, y = 2.

From the second, $x = \dfrac{17 - 4y}{3}$. Put it into the first and multiply by 3:
$$\begin{aligned} 5(17 - 4y) + 18y &= 81 \\ 85 - 2y &= 81 \\ y &= 2 \end{aligned}$$
$x = \dfrac{17 - 8}{3} = 3$. x = 3, y = 2.

From the second, x = 11 − 3y. Then
$$\begin{aligned} 3(11 - 3y) - 2y &= 11 \\ 33 - 11y &= 11 \\ y &= 2 \end{aligned}$$
x = 11 − 6 = 5. x = 5, y = 2.

From the second, x = 8 − 2y. Then
$$\begin{aligned} 3(8 - 2y) - 2y &= 8 \\ 24 - 8y &= 8 \\ y &= 2 \end{aligned}$$
x = 8 − 4 = 4. x = 4, y = 2.

From the second, $y = \dfrac{12 - 3x}{2}$. Put it into the first and multiply by 2:
$$\begin{aligned} 8x - 3(12 - 3x) + 2 &= 0 \\ 17x - 34 &= 0 \\ x &= 2 \end{aligned}$$
$y = \dfrac{12 - 6}{2} = 3$. x = 2, y = 3.

From the first, y = 5x + 1. Then
$$\begin{aligned} 2x - 5(5x + 1) + 51 &= 0 \\ -23x + 46 &= 0 \\ x &= 2 \end{aligned}$$
y = 10 + 1 = 11. x = 2, y = 11.

From the second, $x = \dfrac{17 - 3y}{2}$. Put it into the first and multiply by 2:
$$\begin{aligned} 9(17 - 3y) - 16y &= 24 \\ 153 - 43y &= 24 \\ y &= 3 \end{aligned}$$
$x = \dfrac{17 - 9}{2} = 4$. x = 4, y = 3.

$$\begin{aligned} 3x - 2(5x - 23) &= 4 \\ -7x + 46 &= 4 \\ x &= 6 \end{aligned}$$
y = 30 − 23 = 7. x = 6, y = 7.

From the first, y = 2x − 7. Then x + 2x − 7 = 5, 3x = 12, x = 4; y = 8 − 7 = 1. x = 4, y = 1.

From the first, $y = \dfrac{15 - 3x}{2}$. Put it into the second and multiply by 2:
$$\begin{aligned} 10x - 3(15 - 3x) - 50 &= 0 \\ 19x - 95 &= 0 \\ x &= 5 \end{aligned}$$
$y = \dfrac{15 - 15}{2} = 0$. x = 5, y = 0.

Add: 2x = 12, x = 6. Then 6 + y = 16, y = 10. x = 6, y = 10.
Check: 6 + 10 = 16 ✓; 6 − 10 = −4 ✓.

Multiply the second by 2: 10x − 2y = 46. Subtract the first: 7x = 42, x = 6. Then 30 − y = 23, y = 7. x = 6, y = 7.
Check: 18 − 14 = 4 ✓; 30 − 7 = 23 ✓.

First × 3: 15x − 6y = 6; second × 2: 4x + 6y = 32. Add: 19x = 38, x = 2. Then 4 + 3y = 16, y = 4. x = 2, y = 4.
Check: 10 − 8 = 2 ✓; 4 + 12 = 16 ✓.

Write the first as 2x + 5y = −7. Subtract the second: 7y = −21, y = −3. Then 2x + 6 = 14, x = 4. x = 4, y = −3.
Check: 8 − 15 + 7 = 0 ✓; 8 + 6 = 14 ✓.

First × 3: 27x − 24y = 36; second × 8: 16x + 24y = 136. Add: 43x = 172, x = 4. Then 8 + 3y = 17, y = 3. x = 4, y = 3 (the same pair as 1g, by the other method).
Check: 36 − 24 = 12 ✓; 8 + 9 = 17 ✓.

First × 3: 9x + 12y = 51; second × 2: 10x + 12y = 54. Subtract: x = 3. Then 9 + 4y = 17, y = 2. x = 3, y = 2.
Check: 9 + 8 = 17 ✓; 15 + 12 = 27 ✓.

First × 15: 105x + 120y = −15; second × 8: 80x + 120y = −40. Subtract: 25x = 25, x = 1. Then 7 + 8y = −1, y = −1. x = 1, y = −1.
Check: 7 − 8 = −1 ✓; 10 − 15 = −5 ✓.

Rearrange: 4x − 3y = 16 and 3x + 5y = 12. First × 5: 20x − 15y = 80; second × 3: 9x + 15y = 36. Add: 29x = 116, x = 4. Then 12 + 5y = 12, y = 0. x = 4, y = 0.
Check: 16 − 16 = 0 = 3 × 0 ✓; 5 × 0 = 12 − 12 ✓.

Divide the first by 3: x − y = −2. Divide the second by 2 and rearrange: −x + 2y = 1. Add: y = −1. Then x + 1 = −2, x = −3. x = −3, y = −1.
Check: −9 + 3 + 6 = 0 ✓; −4 − 2 + 6 = 0 ✓.

Rearrange: 3x − 4y = 18 and 5x − 7y = 31. First × 5: 15x − 20y = 90; second × 3: 15x − 21y = 93. Subtract the second from the first: y = −3. Then 3x + 12 = 18, x = 2. x = 2, y = −3.
Check: 6 = −12 + 18 ✓; 10 = −21 + 31 ✓.

Write the second as 5x + 3y = −4. First × 3: 9x + 12y = 6; second × 4: 20x + 12y = −16. Subtract: 11x = −22, x = −2. Then −6 + 4y = 2, y = 2. x = −2, y = 2.
Check: −10 + 6 + 4 = 0 ✓.

Write the second as 8x − 9y = −100. First × 4: 8x + 20y = 480. Subtract the second: 29y = 580, y = 20. Then 2x + 100 = 120, x = 10. x = 10, y = 20.
Check: 80 − 180 + 100 = 0 ✓.

Clear the fractions first: multiply the first by 5 and the second by 3.

6x + 7y = 5 and 7x + 8y = 6. First × 8: 48x + 56y = 40; second × 7: 49x + 56y = 42. Subtract: x = 2. Then 12 + 7y = 5, y = −1. x = 2, y = −1.
Check: $\frac{12}{5} - \frac{7}{5} = 1$ ✓; $\frac{14}{3} - \frac{8}{3} = 2$ ✓.

Multiply the first by 6 and the second by 24, the LCM of 8 and 3.

3x + 2y = 36 and 9x + 24 = 16y, that is 9x − 16y = −24. First × 3: 9x + 6y = 108. Subtract the second: 22y = 132, y = 6. Then 3x + 12 = 36, x = 8. x = 8, y = 6.
Check: 4 + 2 = 6 ✓; $3 + 1 = 4 = \frac{12}{3}$ ✓.

The sum is x + y; three times it, 3(x + y); one more: 3(x + y) + 1.

Anisha (x + 10) years and Tashi (y + 10) years.

(15 − m) years and (11 − m) years. For example, 10 years ago they were 5 and 1.

10x + y, and reversed, 10y + x. For example, with x = 4 and y = 7: 47 and 74.

Let the numbers be x and y, x > y: x + y = 128 and x − y = 16. Add: 2x = 144, x = 72; then y = 56.
The numbers are 72 and 56. Check: 72 + 56 = 128, 72 − 56 = 16 ✓.

x = 5y and x + y = 72. So 6y = 72, y = 12, x = 60. The numbers are 12 and 60.

x + y = 105 and x − y = 15. Add: 2x = 120, x = 60; y = 45.
The angles are 60∘° and 45∘° (and the third angle is $180° - 105° = 75°$).

Let Bishal have x and Sumita y at the start: x = y …(i) and x − 5 = 2(y − 12) − 2 …(ii).
(ii) gives x − 5 = 2y − 26, so x = 2y − 21. With (i): y = 2y − 21, y = 21.
Each had 21 chocolates. Check: Bishal has 16 left, Sumita 9, and 2 × 9 − 2 = 16 ✓.

Let the father be x and the son y years old now: x + y = 46 …(i) and x − 5 = 5(y − 5) …(ii).
(ii) gives x = 5y − 20. Into (i): 6y − 20 = 46, y = 11, x = 35.
The father is 35 years old and the son 11. Check: five years ago, 30 and 6 ✓.

Let the father be x and the daughter y: x − 15 = 4(y − 15) …(i) and x + 5 = 2(y + 5) …(ii).
From (i), x = 4y − 45.
From (ii), x = 2y + 5.
So 4y − 45 = 2y + 5, y = 25, x = 55.
The father is 55 years old and the daughter 25. Check: 15 years ago, 40 and 10 ✓; in 5 years, 60 and 30 ✓.

Let the mother be x and the son y: x + 1 = 4(y + 1) …(i) and x − 2 = 3(y + 4) …(ii).
From (i), x = 4y + 3.
From (ii), x = 3y + 14.
So 4y + 3 = 3y + 14, y = 11, x = 47.
The mother is 47 years old and her son 11. Check: next year 48 and 12 ✓; two years ago she was 45, and in four years he will be 15 ✓.

Let Rajan be x and Nimesh y years old now.
$\dfrac{x - 3}{y - 3} = \dfrac{4}{3}$ gives 3x − 9 = 4y − 12, so 3x = 4y − 3 …(i).
$\dfrac{x + 3}{y + 3} = \dfrac{11}{9}$ gives 9x + 27 = 11y + 33, so 9x = 11y + 6 …(ii).
(i) × 3: 9x = 12y − 9. Subtract (ii): 0 = y − 15, y = 15; then 3x = 57, x = 19.
Rajan is 19 years old and Nimesh 15. Check: three years ago 16 : 12 = 4 : 3 ✓; in three years 22 : 18 = 11 : 9 ✓.

Let the father be x and the son y years old now.
$\dfrac{x - 10}{y - 10} = \dfrac{11}{3}$ gives 3x − 30 = 11y − 110, so 3x = 11y − 80 …(i).
x + 5 = 2(y + 5) + 10 gives x = 2y + 15 …(ii).
Substitute (ii) into (i): 6y + 45 = 11y − 80, 5y = 125, y = 25; x = 65.
The son is 25 and the father 65, so the son will be 65 in 65 − 25 = 40 years.
In 40 years. Check: ten years ago 55 : 15 = 11 : 3 ✓; in five years 70 = 2 × 30 + 10 ✓.

Let the fraction be $\dfrac{x}{y}$: x + y = 2y − 3, so x = y − 3 …(i); and $x - 1 = \dfrac{y - 1}{2}$, so 2x − 2 = y − 1 …(ii).
Substitute (i): 2y − 6 − 2 = y − 1, y = 7, x = 4.
The fraction is $\dfrac{4}{7}$. Check: 4 + 7 = 11 = 14 − 3 ✓; $\dfrac{3}{6}$: 3 is half of 6 ✓.

Let the tens digit be x and the ones digit y.
10x + y = 3(x + y) gives 7x = 2y …(i).
10x + y + 45 = 10y + x gives 9y − 9x = 45, so y = x + 5 …(ii).
Substitute (ii) into (i): 7x = 2x + 10, x = 2, y = 7.
The number is 27. Check: 3 × (2 + 7) = 27 ✓; 27 + 45 = 72 ✓.

Let the tens digit be x and the ones digit y.
(10x + y) + (10y + x) = 121 gives 11(x + y) = 121, so x + y = 11.
The number is the larger, so (10x + y) − (10y + x) = 27 gives 9(x − y) = 27, so x − y = 3.
Add: 2x = 14, x = 7, y = 4. The number is 74. Check: 74 + 47 = 121 ✓, 74 − 47 = 27 ✓, and 74 is larger than 47 ✓.

Let the tens digit be x and the ones digit y: x + y = 9 …(i) and 7(10x + y) = 4(10y + x) …(ii).
(ii) gives 70x + 7y = 40y + 4x, 66x = 33y, so y = 2x. With (i): 3x = 9, x = 3, y = 6. The number is 36 (check: 7 × 36 = 252 = 4 × 63 ✓).
The ones digit, 6, is 100% more than the tens digit, 3 ($\frac{6 - 3}{3} \times 100\%$); the tens digit is 50% less than the ones digit ($\frac{6 - 3}{6} \times 100\%$).

Let onions cost Rs x a kg and sugar Rs y: 5x + 7y = 810 …(i) and 5x = 2y …(ii).
Put (ii) into (i): 2y + 7y = 810, y = 90; then 5x = 180, x = 36.
Onions cost Rs 36 a kg and sugar Rs 90 a kg. Check: 180 + 630 = 810 ✓.

The perimeter is 75000 ÷ 100 = 750 m, so 2(l + b) = 750 and l + b = 375 …(i). The ratio gives 4l = 11b …(ii).
From (i), l = 375 − b; into (ii): 1500 − 4b = 11b, 15b = 1500, b = 100, l = 275.
The land is 275 m long and 100 m wide. Check: 275:100 = 11:4 ✓; 2 × 375 = 750 m, Rs 75,000 ✓.

$b = \dfrac{l}{3}$ …(i) and 2(l + b) = 32, so l + b = 16 …(ii). Substitute: $l + \dfrac{l}{3} = 16$, $\dfrac{4l}{3} = 16$, l = 12, b = 4.
The playground is 12 m long and 4 m wide.

Let her get x notes of Rs 50 and y of Rs 100: x + y = 25 …(i) and 50x + 100y = 2000, so x + 2y = 40 …(ii).
Subtract (i) from (ii): y = 15; x = 10.
10 notes of Rs 50 and 15 notes of Rs 100. Check: 500 + 1500 = 2000 ✓.