Solve 5x + 2y = 15; - 2x + y = 4 | 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

$$5x+2y=15; -2x+y=4$$

Solve for x, y

$x=\frac{7}{9}\approx 0.777777778$
$y = \frac{50}{9} = 5\frac{5}{9} \approx 5.555555556$

Steps Using Substitution

Steps Using Matrices

Steps Using Elimination

Steps Using Substitution

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.

$$5x+2y=15,-2x+y=4$$

Choose one of the equations and solve it for $x$ by isolating $x$ on the left hand side of the equal sign.

$$5x+2y=15$$

Subtract $2y$ from both sides of the equation.

$$5x=-2y+15$$

Divide both sides by $5$.

$$x=\frac{1}{5}\left(-2y+15\right)$$

Multiply $\frac{1}{5}$ times $-2y+15$.

$$x=-\frac{2}{5}y+3$$

Substitute $-\frac{2y}{5}+3$ for $x$ in the other equation, $-2x+y=4$.

$$-2\left(-\frac{2}{5}y+3\right)+y=4$$

Multiply $-2$ times $-\frac{2y}{5}+3$.

$$\frac{4}{5}y-6+y=4$$

Add $\frac{4y}{5}$ to $y$.

$$\frac{9}{5}y-6=4$$

Add $6$ to both sides of the equation.

$$\frac{9}{5}y=10$$

Divide both sides of the equation by $\frac{9}{5}$, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{50}{9}$$

Substitute $\frac{50}{9}$ for $y$ in $x=-\frac{2}{5}y+3$. Because the resulting equation contains only one variable, you can solve for $x$ directly.

$$x=-\frac{2}{5}\times \frac{50}{9}+3$$

Multiply $-\frac{2}{5}$ times $\frac{50}{9}$ by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{20}{9}+3$$

Add $3$ to $-\frac{20}{9}$.

$$x=\frac{7}{9}$$

The system is now solved.

$$x=\frac{7}{9},y=\frac{50}{9}$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$5x+2y=15,-2x+y=4$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&2\\-2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\4\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of $\left(\begin{matrix}5&2\\-2&1\end{matrix}\right)$.

$$inverse(\left(\begin{matrix}5&2\\-2&1\end{matrix}\right))\left(\begin{matrix}5&2\\-2&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\-2&1\end{matrix}\right))\left(\begin{matrix}15\\4\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\-2&1\end{matrix}\right))\left(\begin{matrix}15\\4\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&2\\-2&1\end{matrix}\right))\left(\begin{matrix}15\\4\end{matrix}\right)$$

For the $2\times 2$ matrix $\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$, the inverse matrix is $\left(\begin{matrix}\frac{d}{ad-bc}&\frac{-b}{ad-bc}\\\frac{-c}{ad-bc}&\frac{a}{ad-bc}\end{matrix}\right)$, so the matrix equation can be rewritten as a matrix multiplication problem.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5-2\left(-2\right)}&-\frac{2}{5-2\left(-2\right)}\\-\frac{-2}{5-2\left(-2\right)}&\frac{5}{5-2\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}15\\4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}&-\frac{2}{9}\\\frac{2}{9}&\frac{5}{9}\end{matrix}\right)\left(\begin{matrix}15\\4\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{9}\times 15-\frac{2}{9}\times 4\\\frac{2}{9}\times 15+\frac{5}{9}\times 4\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{9}\\\frac{50}{9}\end{matrix}\right)$$

Extract the matrix elements $x$ and $y$.

$$x=\frac{7}{9},y=\frac{50}{9}$$

Steps Using Elimination

In order to solve by elimination, coefficients of one of the variables must be the same in both equations so that the variable will cancel out when one equation is subtracted from the other.

$$5x+2y=15,-2x+y=4$$

To make $5x$ and $-2x$ equal, multiply all terms on each side of the first equation by $-2$ and all terms on each side of the second by $5$.

$$-2\times 5x-2\times 2y=-2\times 15,5\left(-2\right)x+5y=5\times 4$$

Simplify.

$$-10x-4y=-30,-10x+5y=20$$

Subtract $-10x+5y=20$ from $-10x-4y=-30$ by subtracting like terms on each side of the equal sign.

$$-10x+10x-4y-5y=-30-20$$

Add $-10x$ to $10x$. Terms $-10x$ and $10x$ cancel out, leaving an equation with only one variable that can be solved.

$$-4y-5y=-30-20$$

Add $-4y$ to $-5y$.

$$-9y=-30-20$$

Add $-30$ to $-20$.

$$-9y=-50$$

Divide both sides by $-9$.

$$y=\frac{50}{9}$$

Substitute $\frac{50}{9}$ for $y$ in $-2x+y=4$. Because the resulting equation contains only one variable, you can solve for $x$ directly.

$$-2x+\frac{50}{9}=4$$

Subtract $\frac{50}{9}$ from both sides of the equation.

$$-2x=-\frac{14}{9}$$

Divide both sides by $-2$.

$$x=\frac{7}{9}$$

The system is now solved.

$$x=\frac{7}{9},y=\frac{50}{9}$$