Question

How do you solve `5( x + 1) - ( 1+ x ) ( 1- x ) - ( 5+ 2x ) = ( x + 3) ^ { 2} - 10`?

Answer 1

How do you solve it? One step at a time. ;-)

The solution is `x=1/2 or x=0`.

Explanation:

`5(x+1)−(1+x)(1−x)−(5+2x)=(x+3)^2−10`

Expand the various brackets: multiple 5 through the first, multiply the second and third by each other, leave the fourth alone, square the fifth (multiply it by itself).

`5x+5−(1 - x^2) −(5+2x)=x^2 +6x + 9 −10`

`5x+5−1 + x^2 − 5-2x =x^2 + 6x + 9 −10`

I did that all pretty quickly: here's how each goes with more steps:

`5(x+1) = 5x+5`

`(1+x)(1-x)=(1+x-x-x^2)=1-x^2`

`(x+3)^2 = (x+3)(x+3) = x^2 + 3x + 3x + 9 = x^2 + 6x+ 9`

Now collect 'like terms' - all the `x^2` terms together, all the `x`, all the numbers.

`5x+5−1 + x^2 − 5-2x =x^2 + 6x + 9 −10`

The `x^2` terms cancel, since there is `+x^2` on both sides. The numbers cancel because there is -1 over all on both sides.

`-9x=0`

Solving this, `x=0`.

Answer 2

You first work away all the parentheses.

Explanation:

`(5x+5)-(1-x^2)-(5+2x)=(x^2+6x+9)-10->`

`5x+5-1+x^2-5-2x=x^2+6x+9-10->`

In this case you see that you can cancel the `x^2`'s.
Now put the `x`'s to the left and the numbers to the right. Don't forget to change the signs:

`5x-2x-(6x)=9-10-(5-1-5)->`

`-3x=19-(-1)->-3x=20->x=-20/3`