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Let us take some examples of Algebra Linear Equations to understand the concept I a better way.

2 (3 + x) + 5 = 10x – 2 (x – 1)

2 (3 + x) + 5 = 10x – 2 (x – 1)

ð 6 + 2x + 5 = 10x – 2x + 2

ð 11 + 2x = 8x + 2

Subtracting 2 from both sides, we get

ð 11 + 2x – 2 = 8x + 2 – 2

ð 9 + 2x = 8x

Subtracting 2x from both sides, we get

ð 9 + 2x – 2x = 8x – 2x

ð 9 = 6x

Dividing both sides by 6, we get

9/6 = 6x/6

x = 3/2

Hence x = 3/2 is the solution of equation 2 (3 + x) + 5 = 10x – 2 (x – 1)

5 (m + 2) -10 = 8m – (3m + 7)

5 (m + 2) -10 = 8m – (6m + 9)

ð 5m + 10 – 10 = 8m – 6m – 9

ð 5m = 2m – 9

Subtracting 2m from both sides, we get

5m – 2m = 2m – 9 – 2m

3m = -9

Dividing both sides by 3, we get

3m/3 = -9/3

m = -3

Hence m -3 is the solution of 5 (m + 2) -10 = 8m – (3m + 7).