Solve {l}{6x+2y=130}{3x=4y}

$$\left\{ \begin{array} { l } { 6 x + 2 y = 130 } \\ { 3 x = 4 y } \end{array} \right.$$

Solve for x, y

$$x = \frac{52}{3} = 17\frac{1}{3} \approx 17.333333333$$

$$y=13$$

Steps Using Substitution

Consider the second equation. Subtract 4y from both sides.

$$3x-4y=0$$

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.

$$6x+2y=130,3x-4y=0$$

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

$$6x+2y=130$$

Subtract 2y from both sides of the equation.

$$6x=-2y+130$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-2y+130\right)$$

Multiply \frac{1}{6} times -2y+130.

$$x=-\frac{1}{3}y+\frac{65}{3}$$

Substitute \frac{-y+65}{3} for x in the other equation, 3x-4y=0.

$$3\left(-\frac{1}{3}y+\frac{65}{3}\right)-4y=0$$

Multiply 3 times \frac{-y+65}{3}.

$$-y+65-4y=0$$

Add -y to -4y.

$$-5y+65=0$$

Subtract 65 from both sides of the equation.

$$-5y=-65$$

Divide both sides by -5.

$$y=13$$

Substitute 13 for y in x=-\frac{1}{3}y+\frac{65}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{3}\times 13+\frac{65}{3}$$

Multiply -\frac{1}{3} times 13.

$$x=\frac{-13+65}{3}$$

Add \frac{65}{3} to -\frac{13}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{52}{3}$$

The system is now solved.

$$x=\frac{52}{3},y=13$$

Steps Using Matrices

Consider the second equation. Subtract 4y from both sides.

$$3x-4y=0$$

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

$$6x+2y=130,3x-4y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&2\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}130\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&2\\3&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&2\\3&-4\end{matrix}\right))\left(\begin{matrix}6&2\\3&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&2\\3&-4\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&2\\3&-4\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&2\\3&-4\end{matrix}\right))\left(\begin{matrix}130\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{6\left(-4\right)-2\times 3}&-\frac{2}{6\left(-4\right)-2\times 3}\\-\frac{3}{6\left(-4\right)-2\times 3}&\frac{6}{6\left(-4\right)-2\times 3}\end{matrix}\right)\left(\begin{matrix}130\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{15}&\frac{1}{15}\\\frac{1}{10}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}130\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{15}\times 130\\\frac{1}{10}\times 130\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{52}{3}\\13\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{52}{3},y=13$$

Steps Using Elimination

Consider the second equation. Subtract 4y from both sides.

$$3x-4y=0$$

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.

$$6x+2y=130,3x-4y=0$$

To make 6x and 3x equal, multiply all terms on each side of the first equation by 3 and all terms on each side of the second by 6.

$$3\times 6x+3\times 2y=3\times 130,6\times 3x+6\left(-4\right)y=0$$

Simplify.

$$18x+6y=390,18x-24y=0$$

Subtract 18x-24y=0 from 18x+6y=390 by subtracting like terms on each side of the equal sign.

$$18x-18x+6y+24y=390$$

Add 18x to -18x. Terms 18x and -18x cancel out, leaving an equation with only one variable that can be solved.

$$6y+24y=390$$

Add 6y to 24y.

$$30y=390$$

Divide both sides by 30.

$$y=13$$

Substitute 13 for y in 3x-4y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$3x-4\times 13=0$$

Multiply -4 times 13.

$$3x-52=0$$

Add 52 to both sides of the equation.

$$3x=52$$

Divide both sides by 3.

$$x=\frac{52}{3}$$

The system is now solved.

$$x=\frac{52}{3},y=13$$