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# Solve for y

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5x2-x= 8y2+9x / 4y

Jun 27, 2018

#1
+2339
+2

Solving for a variable can be extremely daunting--especially in a multivariable equation such as $$5x^2-x=8y^2+\frac{9xy}{4}$$. I have a few suggestions that may make this easier to do.

1. Move everything to One Side of the Equation.

This is a relatively simple step.

$$5x^2-x=8y^2+\frac{9xy}{4}\Rightarrow8y^2+\frac{9x}{4}y-5x^2+x=0$$

2. Eliminate All Instances of Fractions or Decimals

Fractions can be pesky, and there is no reason to make a hard situation worse. In this case, we can multiply both sides of the equation by 4 to eliminate the fractions. In a situation like this one, this is also relatively easy to do.

$$8y^2+\frac{9x}{4}y-5x^2+x=0\Rightarrow32y^2+9xy-20x^2+4x=0$$

3. Use a Formula to Finish it Off

This is written in the form of a quadratic, so the quadratic formula is the way to go.

 $$a=32; b=9x;c=-20x^2+4x\\ y_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The only thing left to do is plug in the numbers. $$y_{1,2}=\frac{-9x\pm\sqrt{(9x)^2-4*32(-20x^2+4x)}}{2*32}$$ It is time to simplify. $$y_{1,2}=\frac{-9x\pm\sqrt{81x^2-128(-20x^2+4x)}}{64}$$ $$y_{1,2}=\frac{-9x\pm\sqrt{81x^2+2560x^2-512x}}{64}$$ $$y_{1,2}=\frac{-9x\pm\sqrt{2641x^2-512x}}{64}$$ I have now successfully solve for y.

Jun 27, 2018

#1
+2339
+2

Solving for a variable can be extremely daunting--especially in a multivariable equation such as $$5x^2-x=8y^2+\frac{9xy}{4}$$. I have a few suggestions that may make this easier to do.

1. Move everything to One Side of the Equation.

This is a relatively simple step.

$$5x^2-x=8y^2+\frac{9xy}{4}\Rightarrow8y^2+\frac{9x}{4}y-5x^2+x=0$$

2. Eliminate All Instances of Fractions or Decimals

Fractions can be pesky, and there is no reason to make a hard situation worse. In this case, we can multiply both sides of the equation by 4 to eliminate the fractions. In a situation like this one, this is also relatively easy to do.

$$8y^2+\frac{9x}{4}y-5x^2+x=0\Rightarrow32y^2+9xy-20x^2+4x=0$$

3. Use a Formula to Finish it Off

This is written in the form of a quadratic, so the quadratic formula is the way to go.

 $$a=32; b=9x;c=-20x^2+4x\\ y_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ The only thing left to do is plug in the numbers. $$y_{1,2}=\frac{-9x\pm\sqrt{(9x)^2-4*32(-20x^2+4x)}}{2*32}$$ It is time to simplify. $$y_{1,2}=\frac{-9x\pm\sqrt{81x^2-128(-20x^2+4x)}}{64}$$ $$y_{1,2}=\frac{-9x\pm\sqrt{81x^2+2560x^2-512x}}{64}$$ $$y_{1,2}=\frac{-9x\pm\sqrt{2641x^2-512x}}{64}$$ I have now successfully solve for y.

TheXSquaredFactor Jun 27, 2018