Solve {[(x)^(2)+(y)^(2)=5],[x-y=1]:} | 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

$$\left\{\begin{matrix} { x }^{ 2 } + { y }^{ 2 } =5 \\ x-y=1 \end{matrix}\right.$$

Solve for x, y

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

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.

$$x-y=1,y^{2}+x^{2}=5$$

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, $y^{2}+x^{2}=5$.

$$y^{2}+\left(y+1\right)^{2}=5$$

Square $y+1$.

$$y^{2}+y^{2}+2y+1=5$$

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

$$2y^{2}+2y+1=5$$

Subtract $5$ from both sides of the equation.

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

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

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

Square $1\times 1\times 1\times 2$.

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

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

$$y=\frac{-2±\sqrt{4-8\left(-4\right)}}{2\times 2}$$

Multiply $-8$ times $-4$.

$$y=\frac{-2±\sqrt{4+32}}{2\times 2}$$

Add $4$ to $32$.

$$y=\frac{-2±\sqrt{36}}{2\times 2}$$

Take the square root of $36$.

$$y=\frac{-2±6}{2\times 2}$$

Multiply $2$ times $1+1\times 1^{2}$.

$$y=\frac{-2±6}{4}$$

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

$$y=\frac{4}{4}$$

Divide $4$ by $4$.

$$y=1$$

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

$$y=-\frac{8}{4}$$

Divide $-8$ by $4$.

$$y=-2$$

There are two solutions for $y$: $1$ and $-2$. 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 $-2$ for $y$ in the equation $x=y+1$ and solve to find the corresponding solution for $x$ that satisfies both equations.

$$x=-2+1$$

Add $-2$ to $1$.

$$x=-1$$

The system is now solved.

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