Problems 1-10 : Solve for x in each case.
Problem 1 :
3x + 5 = 19
Problem 2 :
ˣ⁄₃₀ = ²⁄₄₅
Problem 3 :
⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄
Problem 4 :
⁴ˣ⁄₃ - 1 = ¹⁴ˣ⁄₁₅ + ¹⁹⁄₅
Problem 5 :
²⁽ˣ ⁺ ¹⁾⁄₃ = ³⁽ˣ ⁻ ²⁾⁄₅
Problem 6 :
⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11
Problem 7 :
²⁄₍ₓ ₊ ₁₎ = 4 - ˣ⁄₍ₓ ₊ ₁₎
Problem 8 :
⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉ = ⁽ˣ ⁺ ⁷⁾⁄₄
Problem 9 :
(x + 2)(x - 3) + (x + 3)(x - 4) = x(2x - 5)
Problem 10 :
ˣ⁄₀.₁ - ¹⁄₀.₀₁ + ˣ⁄₀.₀₀₁ - ¹⁄₀.₀₀₀₁ = 0
Problem 11 :
Check whether ¼ is a solution to the following linear equation :
3(x + 1) = 3(5 – x) – 2(5 + x)
Problem 12 :
When 5 times a number y is divided by 8, the result is 15. What is the value of y?
Problem 13 :
3 more than 5 times of a number is equal to -12. Find the number.
Problem 14 :
Three times the present age of a man is equal to 10 more than twice his age 5 years hence. Find the present age of the man.
1. Answer :
To find the value of x, we have to isolate it. To isolate x, we have to get rid of all the values around x using the properties of equality.
3x + 5 = 19
Subtract 5 from both sides.
3x = 15
Divide both sides by 3.
x = 5
2. Answer :
ˣ⁄₃₀ = ²⁄₄₅
To solve the above equation, we have to get rid of the denominators 30 and 45.
Least common multiple of (30, 45) = 90.
Multiply both sides of the equation by the least common multiple 15 to get rid of the denominators.
90(ˣ⁄₃₀) = 90(²⁄₄₅)
3x = 2(2)
3x = 4
Divide both sides by 3.
x = ⁴⁄₃
x = 1⅓
3. Answer :
⁽ˣ ⁺ ²⁴⁾⁄₅ = 4 + ˣ⁄₄
Least common multiple of the denominators (5, 4) = 20.
Multiply both sides of the equation by the least common multiple 20 to get rid of the denominators.
20[⁽ˣ ⁺ ²⁴⁾⁄₅] = 20(4 + ˣ⁄₄)
4(x + 24) = 20(4) + 20(ˣ⁄₄)
4x + 96 = 80 + 5x
Subtract 4x from both sides.
96 = 80 + x
Subtract 80 from both sides.
16 = x
4. Answer :
⁴ˣ⁄₃ - 1 = ¹⁴ˣ⁄₁₅ + ¹⁹⁄₅
Least common multiple of the denominators (3, 15, 5) = 15.
Multiply both sides of the equation by the least common multiple 15 to get rid of the denominators.
15(⁴ˣ⁄₃ - 1) = 15(¹⁴ˣ⁄₁₅ + ¹⁹⁄₅)
15(⁴ˣ⁄₃) + 15(-1) = 15(¹⁴ˣ⁄₁₅) + 15(¹⁹⁄₅)
20x - 15 = 14x + 57
Subtract 14x from both sides.
6x - 15 = 57
Add 15 to both sides.
6x = 72
Divide both sides by 6.
x = 12
5. Answer :
²⁽ˣ ⁺ ¹⁾⁄₃ = ³⁽ˣ ⁻ ²⁾⁄₅
Do cross multiplication.
5[2(x + 1)] = 3[3(x - 2)]
10(x + 1) = 9(x - 2)
10x - 10 = 9x - 18
Subtract 9x from both sides.
x - 10 = -18
Add 10 to both sides.
x = -8
6. Answer :
⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃ = 11
Least common multiple of the denominators (4, 3) = 12.
Multiply both sides of the equation by the least common multiple 12 to get rid of the denominators.
12[⁽ˣ ⁺ ⁴⁾⁄₄ + ⁽ˣ ⁻ ⁵⁾⁄₃] = 12(11)
12[⁽ˣ ⁺ ⁴⁾⁄₄] + 12[⁽ˣ ⁻ ⁵⁾⁄₃] = 132
3(x + 4) + 4(x - 5) = 132
3x + 12 + 4x - 20 = 132
7x - 8 = 132
Add 8 to both sides.
7x = 140
Divide both sides by 7.
x = 20
7. Answer :
²⁄₍ₓ ₊ ₁₎ = 4 - ˣ⁄₍ₓ ₊ ₁₎
Multiply both sides by (x + 1) to get rid of the denomintors.
2 = 4(x + 1) - x
Use distributive property.
2 = 4x + 4 - x
2 = 3x + 4
Subtract 4 from both sides.
2 - 4 = 3x
-2 = 3x
Divide both sides by 3.
⁻²⁄₃ = x
8. Answer :
⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉ = ⁽ˣ ⁺ ⁷⁾⁄₄
Least common multiple of the denominators (6, 9, 4) = 36.
Multiply both sides of the equation by the least common multiple 36 to get rid of the denominators.
36[⁽ˣ ⁺ ¹¹⁾⁄₆ - ⁽ˣ ⁺ ¹⁾⁄₉] = 36[⁽ˣ ⁺ ⁷⁾⁄₄]
36[⁽ˣ ⁺ ¹¹⁾⁄₆] - 36[⁽ˣ ⁺ ¹⁾⁄₉] = 36[⁽ˣ ⁺ ⁷⁾⁄₄]
6(x + 11) - 4(x + 1) = 9(x + 7)
6x + 66 - 4x - 4 = 9x + 63
2x + 62 = 9x + 63
Subtract 2x from both sides.
62 = 7x + 63
Subtract 63 from both sides.
-1 = 7x
Divide both sides by 7.
⁻¹⁄₇ = x
9. Answer :
(x + 2)(x - 3) + (x + 3)(x - 4) = x(2x - 5)
x^{2} - 3x + 2x - 6 + x^{2 }- 4x + 3x - 12 = 2x^{2} - 5x
2x^{2} - 2x - 18 = 2x^{2} - 5x
Subtract 2x^{2 }from both sides.
-2x - 18 = -5x
Add 5x to both sides.
3x - 18 = 0
Add 18 to both sides.
3x = 18
Divide both sides by 3.
x = 6
10. Answer :
ˣ⁄₀.₁ - ¹⁄₀.₀₁ + ˣ⁄₀.₀₀₁ - ¹⁄₀.₀₀₀₁ = 0
x(¹⁰⁄₁) - 1(¹⁰⁰⁄₁) + x(¹⁰⁰⁰⁄₁) - 1(¹⁰⁰⁰⁰⁄₁) = 0
10x - 100 + 1000x - 10000 = 0
1010x - 10100 = 0
Add 10100 to both sides.
1010x = 10100
Divide both sides by 1010.
x = 10
11. Answer :
3(x + 1) = 3(5 – x) – 2(5 + x)
To check 1/4 is the solution to the given linear equation, substitute x = 1/4 in the equation.
3(¼ + 1) = 3(5 – ¼) – 2(5 + ¼)
3(⁵⁄₄) = 3(¹⁹⁄₄) - 2(²¹⁄₄)
¹⁵⁄₄ = ⁵⁷⁄₄ - ⁴²⁄₄
¹⁵⁄₄ = ⁽⁵⁷ ⁻ ⁴²⁾⁄₄
¹⁵⁄₄ = ¹⁵⁄₄
Since ¼ makes the given equation true, it is the solution.
12. Answer :
From rthe given information, we have
⁵ʸ⁄₈ = 15
Multiply both sides by 8.
5y = 120
Divide both sides by 5.
y = 24
13. Answer :
Let x be the required number.
It is given that 3 more than 5 times of a number is equal to -12.
5x + 3 = -12
Subtract 3 from both sides.
5x = -15
Divide both sides by 5.
x = -3
The number is -3.
14. Answer :
Let x be the present age of the man.
Age of the man 5 years hence = x + 5
It is given that three times the present age of the man is equal to 10 more than twice his age 5 years hence.
3x = 2(x + 5) + 10
3x = 2x + 10 + 10
3x = 2x + 20
Subtract 2x from both sides.
x = 20
The present age of the man is 20 years.
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Oct 02, 23 11:40 PM
Oct 02, 23 08:32 PM
Oct 02, 23 12:38 PM