Solve =5x^2+34x-7y^2=(5x-4)(x+7y)

$$= 5 x ^ { 2 } + 34 x - 7 y ^ { 2 } = ( 5 x - 4 ) ( x + 7 y )$$

Solve for x

$$x=-\frac{7y\left(4-y\right)}{38-35y}$$

$$y\neq \frac{38}{35}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5x-4 by x+7y.

$$5x^{2}+34x-7y^{2}=5x^{2}+35xy-4x-28y$$

Subtract 5x^{2} from both sides.

$$5x^{2}+34x-7y^{2}-5x^{2}=35xy-4x-28y$$

Combine 5x^{2} and -5x^{2} to get 0.

$$34x-7y^{2}=35xy-4x-28y$$

Subtract 35xy from both sides.

$$34x-7y^{2}-35xy=-4x-28y$$

Add 4x to both sides.

$$34x-7y^{2}-35xy+4x=-28y$$

Combine 34x and 4x to get 38x.

$$38x-7y^{2}-35xy=-28y$$

Add 7y^{2} to both sides.

$$38x-35xy=-28y+7y^{2}$$

Combine all terms containing x.

$$\left(38-35y\right)x=-28y+7y^{2}$$

The equation is in standard form.

$$\left(38-35y\right)x=7y^{2}-28y$$

Divide both sides by -35y+38.

$$\frac{\left(38-35y\right)x}{38-35y}=\frac{7y\left(y-4\right)}{38-35y}$$

Dividing by -35y+38 undoes the multiplication by -35y+38.

$$x=\frac{7y\left(y-4\right)}{38-35y}$$

Solve for y

$$y=-\frac{\sqrt{1225x^{2}-896x+784}}{14}-\frac{5x}{2}+2$$

$$y=\frac{\sqrt{1225x^{2}-896x+784}}{14}-\frac{5x}{2}+2$$