(x-1)(x+2)=5(x-1) divided by x-1

1 answer:

5 0

Assuming that your question is (x-1)(x+2) = [5(x-1)]/x-1

On the right side, the x-1's will cancel out, leaving you with (x-1)(x+2) = 5

expand the left side, giving you x^2 + x -2 = 5
which goes to x^2 +x -7 = 0
the possible values for x are 2.93 and -3.93. I don't think this was your question, so I'll do the other possible question that you might have been asking.

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

divide by x-1 on both sides, leaving you with x+2=5
x+2=5
x=5-3
x=3

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Read 2 more answers

The perimeter of a rectangle is 40cm. If the length were doubled and the width halved, the perimeter would be increased by 16cm.

alexgriva [62]

Answer:

The dimensions of the original rectangle are Length =12cm, Width=8cm

Step-by-step explanation:

Let the length of the rectangle be x

Let the width of the rectangle be y

The perimeter of a rectangle is 40cm.

2(x+y)=40

Divide both sides by 2

x+y=20

If the length were doubled(2x) and the width halved(y/2), the perimeter would be increased by 16cm,i.e.(40+16)cm

Therefore:

$2(2x+\dfrac{y}{2})=56$

Divide both sides by 2

$2x+\dfrac{y}{2}=28$

From the first equation, x=20-y.

Substitute x=20-y into $2x+\dfrac{y}{2}=28$

$2x+\dfrac{y}{2}=28$

$2(20-y)+\dfrac{y}{2}=28\\\dfrac{4(20-y)+y}{2}=28\\80-4y+y=56\\-3y=-24\\$Divide both sides by -3\\y=8cm\\Recall:x=20-y\\x=20-8=12cm\\The dimensions of the original rectangle are Length =12cm, Width=8cm$