What is the answer to solvefor x,p^2x+q^2y=z ?

The answer to solvefor x,p^2x+q^2y=z is x=(z-yq^2)/(p^2);\quad p\ne 0

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solvefor x,p^2x+q^2y=z

solve for

x, p^{2} x + q^{2} y = z

x = \frac{z - y q ^{2}}{p ^{2}}; p \neq = 0

Solution steps

p^{2} x + q^{2} y = z

Move

q^{2} y

to the right side

p^{2} x = z - q^{2} y

Divide both sides by

p^{2}; p \neq = 0

x = \frac{z - y q ^{2}}{p ^{2}}; p \neq = 0

\frac{( 2 x - 5 )}{4} - \frac{( 8 x - 6 )}{5} = \frac{( 6 + 2 )}{7}

m = \frac{( 1 2 ^{7} \cdot 0.1 8 ^{3} \cdot 0.64 )}{( 2 4 ^{8} \cdot 0.81 )}

y, 2 y + 5 = x

- 55 c (c - 2.25) = - 11 c (5 c - 8.7) - 1.5

b, l = 2 h + b

What is the answer to solvefor x,p^2x+q^2y=z ?
The answer to solvefor x,p^2x+q^2y=z is x=(z-yq^2)/(p^2);\quad p\ne 0