SOLVING LINEAR EQUATIONS WITH FRACTIONS

Problem 1 :

(x + 1) / 4 = (x - 2) / 3

Solution :

(x + 1) / 4 = (x - 2) / 3

Doing cross multiplication, we get

3(x + 1) = 4(x - 2)

Using distributive property, we get

3x + 3 = 4x - 8

Subtract 4x on both sides.

3x - 4x + 3 = -8

-x + 3 = -8

Subtract 3 on both sides.

-x = -8 - 3

-x = -11

Divide by -1 on both sides.

x = 11

Problem 2 :

(2x - 1) / 5 = (3x + 1) / 3

Solution :

(2x - 1) / 5 = (3x + 1) / 3

On both sides of the equal sign, we have fractions. By doing cross multiplication, we get

3(2x - 1) = 5(3x + 1)

Using distributive property, we get

6x - 3 = 15x + 5

Subtracting 15x on both sides and add 3 on both sides, we get.

6x - 15x = 5 + 3

-9x = 8

x = -8/9

Problem 3 :

3x - (x - 2)/3 = 4 - (x - 1)/4

Solution :

3x - (x - 2)/3 = 4 - (x - 1)/4

(9x - x + 2) / 3= (16 - x + 1) / 4

(8x + 2) / 3 = (17 - x) / 4

Cross multiplying both sides.

4(8x + 2) = 3(17 - x)

32x + 8 = 51 - 3x

32x + 3x = 51 -8 

35x = 43

x = 43/35

So, the answer is x = 43/35.

Problem 4 :

(3t + 5/4) - 1 = (4t - 3) / 5

Solution :

In LHS,LCM is 4.

(3t + 5 - 4) / 4= (4t - 3) / 5

(3t + 1) / 4 = (4t - 3) / 5

Doing cross multiplication.

5(3t + 1) = 4(4t - 3)

15t + 5 = 16t - 12

15t - 16t = -12 - 5

-t = -17

t = 17

Hence value of t is 17.

Problem 5 :

(2y - 3)/4 - (3y - 5)/2 = y + 3/4

Solution :

Find common denominator.

LCM of 4, 2 is 4.

[(2y - 3) - 2(3y - 5)] / 4= y + 3/4

By distributing

(2y - 3 - 6y + 10) / 4 = y + 3/4

Combine like terms.

(-4y + 7) / 4 = y + 3/4

Subtract y from both sides.

(-4y + 7)/4 - y = y + 3/4-y

(-4y + 7 - 4y) / 4 = 3/4

(-8y + 7) / 4 = 3/4

Multiplying by 4 on both sides, we get

-8y + 7 = 3

-8y = 3 - 7

-8y = -4

y = 1/2

So, the answer is y = 1/2.

Problem 6 :

(3t - 2) / 3 + (2t + 3) / 2 = t + 7/6

Solution :

Find common denominator.

LCM of 3, 2 is 6.

(3t - 2) / 3 + (2t + 3) / 2 = t + 7/6

Least common multiple for 2 and 3 is 6.

[2(3t - 2) + 3(2t + 3)] / 6 = t + 7/6

[(6t -  4)+(6t + 9)] / 6 = t + 7/6

(12t+5) / 6 = t + 7/6

Subtract t from both sides.

((12t+5)/6) - t = t + 7/6 - t

(12t + 5 - 6t)/6 = 7/6

(6t + 5) / 6 = 7/6

Multiplying by 6 on both sides, we get

6t + 5 = 7

6t = 7 - 5

6t = 2

t = 1/3

So, the answer is t = 1/3.

Problem 7 :

m - (m - 1)/2 = 1 - (m - 2) / 3

Solution :

[m - (m - 1)/2 = 1 - (m - 2) / 3

(2m - m + 1) / 2 = (3 - m + 2) / 3

Simplify like terms.

(m + 1) / 2 = (5 - m) / 3

Doing cross multiplication, we get

3(m + 1) = 2(5 - m)

3m + 3 = 10 - 2m

3m + 2m = 10 - 3

5m = 7

m = 7/5

So, the answer is m = 7/5.