Solve x ^ 2 + y ^ 2 = 34; x - y = 2 | Equatix - Math Solver

Fractions & Decimals

Ratios & Proportions

Basic Number Theory

Integers & Absolute Values

Factors & Multiples

Prime Factorization

Basic Equations & Inequalities

Coordinate Plane & Graphing

Linear Equations & Inequalities

Quadratic Equations

Polynomials & Factoring

Functions & Graphs

Systems of Equations

Trigonometric Functions

Trigonometric Identities & Equations

Right Triangle Trigonometry

Graphing Trigonometric Functions

Limits & Continuity

Derivatives & Applications

Integrals & Applications

Series & Sequences

Multivariable Calculus

Functions & Graphs

Polynomial & Rational Functions

Exponential & Logarithmic Functions

Sequences & Series

Descriptive Statistics

Normal Distribution

Hypothesis Testing

Regression Analysis

Linear Programming

Sets & Counting

Eigenvalues & Eigenvectors

Linear Transformations

Atomic Structure

Thermochemistry

Example

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

a⋅(b-2)=3b

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

a⋅(b-2)=3b

Question

$$x^{2}+y^{2}=34; x-y=2$$

Solve for x, y

$x=5\text{, }y=3$
$x=-3\text{, }y=-5$

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.

$$x-y=2,y^{2}+x^{2}=34$$

Solve $x-y=2$ for $x$ by isolating $x$ on the left hand side of the equal sign.

$$x-y=2$$

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

$$x=y+2$$

Substitute $y+2$ for $x$ in the other equation, $y^{2}+x^{2}=34$.

$$y^{2}+\left(y+2\right)^{2}=34$$

Square $y+2$.

$$y^{2}+y^{2}+4y+4=34$$

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

$$2y^{2}+4y+4=34$$

Subtract $34$ from both sides of the equation.

$$2y^{2}+4y-30=0$$

This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1+1\times 1^{2}$ for $a$, $1\times 2\times 1\times 2$ for $b$, and $-30$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.

$$y=\frac{-4±\sqrt{4^{2}-4\times 2\left(-30\right)}}{2\times 2}$$

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

$$y=\frac{-4±\sqrt{16-4\times 2\left(-30\right)}}{2\times 2}$$

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

$$y=\frac{-4±\sqrt{16-8\left(-30\right)}}{2\times 2}$$

Multiply $-8$ times $-30$.

$$y=\frac{-4±\sqrt{16+240}}{2\times 2}$$

Add $16$ to $240$.

$$y=\frac{-4±\sqrt{256}}{2\times 2}$$

Take the square root of $256$.

$$y=\frac{-4±16}{2\times 2}$$

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

$$y=\frac{-4±16}{4}$$

Now solve the equation $y=\frac{-4±16}{4}$ when $±$ is plus. Add $-4$ to $16$.

$$y=\frac{12}{4}$$

Divide $12$ by $4$.

$$y=3$$

Now solve the equation $y=\frac{-4±16}{4}$ when $±$ is minus. Subtract $16$ from $-4$.

$$y=-\frac{20}{4}$$

Divide $-20$ by $4$.

$$y=-5$$

There are two solutions for $y$: $3$ and $-5$. Substitute $3$ for $y$ in the equation $x=y+2$ to find the corresponding solution for $x$ that satisfies both equations.

$$x=3+2$$

Add $1\times 3$ to $2$.

$$x=5$$

Now substitute $-5$ for $y$ in the equation $x=y+2$ and solve to find the corresponding solution for $x$ that satisfies both equations.

$$x=-5+2$$

Add $-5$ to $2$.

$$x=-3$$

The system is now solved.

$$x=5,y=3\text{ or }x=-3,y=-5$$