Problem 1 :
Solve for x :
⁴⁄₃ₓ + ⅓ = 1
Problem 2 :
Solve for a :
⁵ᵃ⁄₍ₐ ₊ ₄₎ = ⁵⁄₂
Problem 3 :
Solve for x :
⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎ = ⁵⁄₃
Problem 4 :
Solve for m :
⅙ + ¹⁄₁₂ = ¹⁄m
Problem 5 :
Solve for x :
ˣ⁄₍ₓ ₊ ₃₎ = ⁸⁄₍ₓ ₊ ₆₎
Problem 6 :
Solve for y :
⁽ʸ ⁺ ³⁾⁄₂y = ⅔
Problem 7 :
Solve for s :
⁹⁄₍₂s ₊ ₇₎ = ³⁄s
Problem 8 :
Solve for y :
⁽⁴ʸ ⁻ ¹⁾⁄₅y = ³⁄y
Problem 9 :
Solve for w :
⁴⁄₍₃w ₊ ₇₎ = ½
Problem 10 :
Solve for y :
¹²⁄y = ⁹⁄₍y ₋ ₃₎
Problem 11 :
Solve for x :
⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎ = ³⁄₂
Problem 12 :
Solve for y :
⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎
1. Answer :
⁴⁄₃ₓ + ⅓ = 1
Multiply both sides of the equation by 3x to get rid of the denominators 3x and 3.
3x(⁴⁄₃ₓ + ⅓) = 1(3x)
3x(⁴⁄₃ₓ) + 3x(⅓) = 3x
4 + x = 3x
Subtract x from both sides.
4 = 2x
Divide both sides by 2.
2 = x
2. Answer :
⁵ᵃ⁄₍ₐ ₊ ₄₎ = ⁵⁄₂
Multiply both sides of the equation by (a + 4) to get rid of the denominator (a + 4) on the left side.
(a + 4)[⁵ᵃ⁄₍ₐ ₊ ₄₎] = (a + 4)(⁵⁄₂)
5a = ⁵⁽ᵃ ⁺ ⁴⁾⁄₂
Multiply both sides by 2.
2(5a) = 5(a + 4)
10a = 5(a + 4)
10a = 5a + 20
Subtract 5a from both sides.
5a = 20
Divide both sides by 5.
a = 4
3. Answer :
⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎ = ⁵⁄₃
Multiply both sides of the equation by (x + 4) to get rid of the denominator (x + 4) on the left side.
(x + 4)[⁽³ˣ ⁺ ¹²⁾⁄₍ₓ ₊ ₄₎] = (x + 4)(⁵⁄₃)
3x + 12 = ⁵⁽ˣ ⁺ ⁴⁾⁄₃
Multiply both sides by 3.
3(3x + 12) = 5(x + 4)
9x + 36 = 5x + 20
Subtract 5x from both sides.
4x + 36 = 20
Subtract 36 from both sides.
4x = -16
Divide both sides by 4.
x = -4
4. Answer :
⅙ + ¹⁄₁₂ = ¹⁄m
Least common multiple of (6, 12) = 12.
Multiply both sides of the equation by 12 to get rid of the denominators 6 and 12 on the left side.
12(⅙ + ¹⁄₁₂) = 12(¹⁄m)
12(⅙) + 12(¹⁄₁₂) = ¹²⁄m
2 + 1 = ¹²⁄m
3 = ¹²⁄m
Multiply both sides by m.
3m = 12
Divide both sides of the equation by 3.
m = 4
5. Answer :
ˣ⁄₍ₓ ₊ ₃₎ = ⁸⁄₍ₓ ₊ ₆₎
Multiply both sides of the equation by (x + 3) to get rid of the denominator (x + 3) on the left side.
(x + 3)[ˣ⁄₍ₓ ₊ ₃₎)] = (x + 3)[⁸⁄₍ₓ ₊ ₆₎]
x = ⁸⁽ˣ ⁺ ³⁾⁄₍ₓ ₊ ₆₎
x = ⁽⁸ˣ ⁺ ²⁴⁾⁄₍ₓ ₊ ₆₎
Multiply both sides by (x + 6).
(x + 6)(x) = 8x + 24
x2 + 6x = 8x + 24
Subtract 8x from both sides.
x2 - 2x = 24
Subtract 24 from both sides.
x2 - 2x - 24 = 0
Solve by factoring.
x2 - 6x + 4x - 24 = 0
x(x - 6) + 4(x - 6) = 0
(x - 6)(x + 4) = 0
x - 6 = 0 or x + 4 = 0
x = 6 or x = -4
6. Answer :
⁽ʸ ⁺ ³⁾⁄₂y = ⅔
Multiply both sides of the equation by 2y to get rid of the denominator 2y on the left side.
2y[⁽ʸ ⁺ ³⁾⁄₂y] = (2y)(⅔)
y + 3 = ⁴ʸ⁄₃
Multiply both sides by 3.
3(y + 3) = 4y
3y + 9 = 4y
Subtract 3y from both sides.
9 = y
7. Answer :
⁹⁄₍₂s ₊ ₇₎ = ³⁄s
Multiply both sides of the equation by (2s + 7) to get rid of the denominator (2s + 7) on the left side.
(2s + 7)[⁹⁄₍₂s ₊ ₇₎] = (2s + 7)[³⁄s]
9 = ⁽⁶ˢ ⁺ ²¹⁾⁄s
Multiply both sides by s
9s = 6s + 21
Subtract 6s from both sides.
3s = 21
Divide both sides by 3.
s = 7
8. Answer :
⁽⁴ʸ ⁻ ¹⁾⁄₅y = ³⁄y
Multiply both sides of the equation by 5y to get rid of the denominators 5y and y.
5y[⁽⁴ʸ ⁻ ¹⁾⁄₅y] = 5y(³⁄y)
4y - 1 = 5(3)
4y - 1 = 15
Add 1 to both sides.
4y = 16
Divide both sides by 4.
y = 4
9. Answer :
⁴⁄₍₃w ₊ ₇₎ = ½
Multiply both sides of the equation by (3w + 7) to get rid of the denominator (3w + 7) on the left side.
(3w + 7)[⁴⁄₍₃w ₊ ₇₎] = (3w + 7)(½)
4 = ⁽³ʷ ⁺ ⁷⁾⁄₂
Multiply both sides by 2.
2(4) = 3w + 7
8 = 3w + 7
Subtract 7 from both sides.
1 = 3w
Divide both sides by 3.
⅓ = w
10. Answer :
¹²⁄y = ⁹⁄₍y ₋ ₃₎
Multiply both sides of the equation by y to get rid of the denominator y on the left side.
y(¹²⁄y) = y[⁹⁄₍y ₋ ₃₎]
12 = ⁹ʸ⁄₍y ₋ ₃₎
Multiply both sides by (y - 3).
(y - 3)(12) = -9y
12y - 36 = 9y
Subtract 9y from both sides.
3y - 36 = 0
Add 36 to both sides.
3y = 36
Divide both sides by 3.
y = 12
11. Answer :
⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎ = ³⁄₂
Multiply both sides of the equation by (2 - x) to get rid of the denominators (2 - x) on the right side.
(2 - x)[⁽³ˣ ⁻ ⁶⁾⁄₍₂ ₋ ₓ₎] = (2 - x)(³⁄₂)
3x - 6 = ⁽⁶ ⁻ ³ˣ⁾⁄₂
Multiply both sides by 2.
2(3x - 6) = 6 - 3x
6x - 12 = 6 - 3x
Add 3x to both sides.
9x - 12 = 6
Add 12 to both sides.
9x = 18
Divide both sides by 9.
x = 2
But, x = 2 makes one of the denominators zero in the given rational equation. So, x = 2 is an extraneous solution, hence it can not be considered as a solution to the given rational equation.
Therefore, there is NO SOLUTION for the given rational equation.
12. Answer :
⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎ = ²⁄y + ¹⁄₍y ₊ ₂₎
Multiply both sides of the equation by (y + 2) to get rid of the denominators (y + 2) on both sides.
(y + 2)[⁽ʸ ⁺ ³⁾⁄₍y ₊ ₂₎] = (y + 2)[²⁄y + ¹⁄₍y ₊ ₂₎]
y + 3 = (y + 2)(²⁄y) + (y + 2)[¹⁄₍y ₊ ₂₎]
y + 3 = ⁽²ʸ ⁺ ⁴⁾⁄y + 1
Multiply both sides by y to get rid of the denominator y on the right side.
y(y + 3) = y[⁽²ʸ ⁺ ⁴⁾⁄y + 1]
y(y) + y(3) = y[⁽²ʸ ⁺ ⁴⁾⁄y] + y(1)
y2 + 3y = 2y + 4 + y
y2 + 3y = 3y + 4
Subtract 3y from both sides.
y2 = 4
Take square root on both sides.
√y2 = √4
y = ±2
y = -2 or y = 2
We get two solutions y = 2 and x = -2 for the given rational equation.
But, y = -2 makes two of the denominators zero in the given rational equation. So, y = -2 is an extraneous solution, hence it is removed from the solutions.
Therefore, solution for the given rational equation is
y = 2
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