Solve x - y = 1; 4y ^ 2 - x ^ 2 = 0 | Equatix - Math Solver

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Example

3(r+2s)=2t-4

a⋅(b-2)=3b

3(r+2s)=2t-4

a⋅(b-2)=3b

Question

$$x-y=1; 4y^{2}-x^{2}=0$$

Solve for x, y

$x=2\text{, }y=1$
$x=\frac{2}{3}\approx 0.666666667\text{, }y=-\frac{1}{3}\approx -0.333333333$

Steps Using Substitution and Quadratic Formula

To solve a pair of equations using substitution, first solve one of the equations for one of the variables. Then substitute the result for that variable in the other equation.

$$x-y=1,4y^{2}-x^{2}=0$$

Solve $x-y=1$ for $x$ by isolating $x$ on the left hand side of the equal sign.

$$x-y=1$$

Subtract $-y$ from both sides of the equation.

$$x=y+1$$

Substitute $y+1$ for $x$ in the other equation, $4y^{2}-x^{2}=0$.

$$4y^{2}-\left(y+1\right)^{2}=0$$

Square $y+1$.

$$4y^{2}-\left(y^{2}+2y+1\right)=0$$

Multiply $-1$ times $y^{2}+2y+1$.

$$4y^{2}-y^{2}-2y-1=0$$

Add $4y^{2}$ to $-y^{2}$.

$$3y^{2}-2y-1=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $4-1^{2}$ for $a$, $-2$ for $b$, and $-1$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-1\right)}}{2\times 3}$$

Square $-2$.

$$y=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-1\right)}}{2\times 3}$$

Multiply $-4$ times $4-1^{2}$.

$$y=\frac{-\left(-2\right)±\sqrt{4-12\left(-1\right)}}{2\times 3}$$

Multiply $-12$ times $-1$.

$$y=\frac{-\left(-2\right)±\sqrt{4+12}}{2\times 3}$$

Add $4$ to $12$.

$$y=\frac{-\left(-2\right)±\sqrt{16}}{2\times 3}$$

Take the square root of $16$.

$$y=\frac{-\left(-2\right)±4}{2\times 3}$$

The opposite of $-2$ is $2$.

$$y=\frac{2±4}{2\times 3}$$

Multiply $2$ times $4-1^{2}$.

$$y=\frac{2±4}{6}$$

Now solve the equation $y=\frac{2±4}{6}$ when $±$ is plus. Add $2$ to $4$.

$$y=\frac{6}{6}$$

Divide $6$ by $6$.

$$y=1$$

Now solve the equation $y=\frac{2±4}{6}$ when $±$ is minus. Subtract $4$ from $2$.

$$y=-\frac{2}{6}$$

Reduce the fraction $\frac{-2}{6}$ to lowest terms by extracting and canceling out $2$.

$$y=-\frac{1}{3}$$

There are two solutions for $y$: $1$ and $-\frac{1}{3}$. Substitute $1$ for $y$ in the equation $x=y+1$ to find the corresponding solution for $x$ that satisfies both equations.

$$x=1+1$$

Add $1\times 1$ to $1$.

$$x=2$$

Now substitute $-\frac{1}{3}$ for $y$ in the equation $x=y+1$ and solve to find the corresponding solution for $x$ that satisfies both equations.

$$x=-\frac{1}{3}+1$$

Add $-\frac{1}{3}$ to $1$.

$$x=\frac{2}{3}$$

The system is now solved.

$$x=2,y=1\text{ or }x=\frac{2}{3},y=-\frac{1}{3}$$