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pochemuha
2 years ago
15

For whatvalue of a is 8x-8+3ax=5ax-2a an identity?

Mathematics
2 answers:
Andrej [43]2 years ago
8 0

Answer with Step-by-step explanation:

We have to solve for a

8x - 8 + 3ax = 5ax - 2a

Subtract both sides by 5ax

8x - 8 + 3ax - 5ax = - 2a

Taking terms of x common

(8+3a-5a)x - 8 = -2a

(8-2a)x - 8 = -2a

Adding both sides by 8 , we get

(8-2a)x = 8-2a

For equation to be true for all values of x,  8-2a must be equal to 0

i.e. 8-2a = 0

⇒ 8 = 2a

dividing both sides by 2, we get

a = 4

Hence, for a=4 , 8x-8+3ax=5ax-2a is an identity

gayaneshka [121]2 years ago
6 0
<span>Identity means that equation is true no matter what.

So we just need to rearrange equation.

8x - 8 + 3ax    =    5ax - 2a

(8+3a)x  - 8    =    (5a)x - 2a

Subtract both sides by 5ax

(8+3a-5a)x - 8 = -2a

(8-2a)x - 8 = -2a

Now add both sides by 8

(8-2a)x = 8-2a

Notice that coefficient of x in left side is same as constant in right side.

For equation to be true, no matter what value of x, 8-2a must be 0

Set 8-2a equal to 0

8-2a = 0

8 = 2a

a = 4

Final answer: a = 4</span>
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