What is the answer to solvefor x,(k-3)x-(4-k^2)y+k^2-7k+6=0 ?

The answer to solvefor x,(k-3)x-(4-k^2)y+k^2-7k+6=0 is x=(y(-k^2+4)-k^2+7k-6)/(k-3);\quad k\ne 3

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solvefor x,(k-3)x-(4-k^2)y+k^2-7k+6=0

solve for

x, (k - 3) x - (4 - k^{2}) y + k^{2} - 7 k + 6 = 0

x = \frac{y ( - k ^{2} + 4 ) - k ^{2} + 7 k - 6}{k - 3}; k \neq = 3

Solution steps

(k - 3) x - (4 - k^{2}) y + k^{2} - 7 k + 6 = 0

Move

(4 - k^{2}) y

to the right side

(k - 3) x + k^{2} - 7 k + 6 = y (- k^{2} + 4)

Move

k^{2}

to the right side

(k - 3) x - 7 k + 6 = y (- k^{2} + 4) - k^{2}

Move

7 k

to the right side

(k - 3) x + 6 = y (- k^{2} + 4) - k^{2} + 7 k

Move

6

to the right side

(k - 3) x = y (- k^{2} + 4) - k^{2} + 7 k - 6

Divide both sides by

k - 3; k \neq = 3

x = \frac{y ( - k ^{2} + 4 ) - k ^{2} + 7 k - 6}{k - 3}; k \neq = 3

\frac{2}{3} (2 y - 3) = \frac{1}{2} (4 y + 2)

5 (x + 5) = 2 x + 26 + 2 x

9 - 4 (3 x + 1) = 3 (6 - 3 x) - 9

- (4 m + 2) + 2 m = 16 - 6 (2 m + 3)

x, 4 x - y = 14

What is the answer to solvefor x,(k-3)x-(4-k^2)y+k^2-7k+6=0 ?
The answer to solvefor x,(k-3)x-(4-k^2)y+k^2-7k+6=0 is x=(y(-k^2+4)-k^2+7k-6)/(k-3);\quad k\ne 3