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Q: How do you solve for x 4x 12?

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4x - 7y = 12-7y = -4x + 12y = (4/7)x - (12/7)======================4x - 7y = 124x = 7y + 12x = (7/4)x + 3

Let the number be x: 4x-12 = 60 4x = 60+12 4x = 72 x = 18

4x-17=31 +17 +17 -------------- 4x = 48 --- ---- 4 4 x = 12

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7x = 3x - 12 4x=-12 x=-3

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4x - 7y = 12-7y = -4x + 12y = (4/7)x - (12/7)======================4x - 7y = 124x = 7y + 12x = (7/4)x + 3

4x+2=2x-102x=-12 x=-6

8x+30=4x+18 -4x -4x 4x+30=18 -30 -30 4x=-12 4x/4=-12/4 x=-3

Let the number be x: 4x-12 = 60 4x = 60+12 4x = 72 x = 18

-4x + 8 = -4 -4x + 8 - 8 = -4 -8 -4x = -12 -4x/-4 = -12/-4 x = 3

4x-17=31 +17 +17 -------------- 4x = 48 --- ---- 4 4 x = 12

16

Is there an operator missing from this question? * 3 + 4x = 19 => x = 4 * 3 - 4x = 19 => x = -4 * 3 . 4x = 19 => x = 19/12

7x = 3x - 12 4x=-12 x=-3

-2x = -4x + 24 -2x + 4x = 24 2x = 24 x = 24/2 x = 12

6(X-2)-4X=16before you can solve for x you first have to bring the x terms togetherand to do that you distribute the 6 to remove the parenthesis6x-(6)2-4x=16 turns into6x-12-4x=16because -4x and 6x are terms of the same degree they can be added togetheryoull get2x-12=16now solve for x2x-12=16 get x term alone by addind 122x=28 now divide 2x=14

We have x - y + 6 = 0 and y = x^2 + 4x - 12; so that, x + 6 = x^2 + 4x - 12 and x^2 + 3x - 18 = 0. So on inspection; x = 3 and y = 9 ANS.