What is the answer to 4(x-1)+x^2-ax=(3-x)^2 ?

The answer to 4(x-1)+x^2-ax=(3-x)^2 is x=(13)/(-a+10);\quad a\ne 10

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

4 (x - 1) + x^{2} - a x = (3 - x)^{2}

x = \frac{13}{- a + 10}; a \neq = 10

Solution steps

4 (x - 1) + x^{2} - a x = (3 - x)^{2}

Expand

4 (x - 1) : 4 x - 4

4 x - 4 + x^{2} - a x = (3 - x)^{2}

Expand

(3 - x)^{2} : 9 - 6 x + x^{2}

4 x - 4 + x^{2} - a x = 9 - 6 x + x^{2}

Move

4

to the right side

4 x + x^{2} - a x = x^{2} - 6 x + 13

Subtract

x^{2} - 6 x

from both sides

4 x + x^{2} - a x - (x^{2} - 6 x) = x^{2} - 6 x + 13 - (x^{2} - 6 x)

Simplify

- a x + 10 x = 13

Factor

- a x + 10 x : x (- a + 10)

x (- a + 10) = 13

Divide both sides by

- a + 10; a \neq = 10

x = \frac{13}{- a + 10}; a \neq = 10

5 y = 35

4 x + 25 = 71

- 2 (d + 6) = - 10

12 + 3 (2 x - 5) = 9

7 x + 10 (2 x + 9) = 36

What is the answer to 4(x-1)+x^2-ax=(3-x)^2 ?
The answer to 4(x-1)+x^2-ax=(3-x)^2 is x=(13)/(-a+10);\quad a\ne 10