Solve {l}{3x-4y=8x-5}{4x+8y=3y+13}{(x,y)=}

$$\left. \begin{array}{l}{ 3 x - 4 y = 8 x - 5 }\\{ 4 x + 8 y = 3 y + 13 }\\{ ( x , y ) = }\end{array} \right.$$

Solve for x, y

$$x=-3$$

$$y=5$$

Steps Using Substitution

Consider the first equation. Subtract 8x from both sides.

$$3x-4y-8x=-5$$

Combine 3x and -8x to get -5x.

$$-5x-4y=-5$$

Consider the second equation. Subtract 3y from both sides.

$$4x+8y-3y=13$$

Combine 8y and -3y to get 5y.

$$4x+5y=13$$

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.

$$-5x-4y=-5,4x+5y=13$$

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

$$-5x-4y=-5$$

Add 4y to both sides of the equation.

$$-5x=4y-5$$

Divide both sides by -5.

$$x=-\frac{1}{5}\left(4y-5\right)$$

Multiply -\frac{1}{5} times 4y-5.

$$x=-\frac{4}{5}y+1$$

Substitute -\frac{4y}{5}+1 for x in the other equation, 4x+5y=13.

$$4\left(-\frac{4}{5}y+1\right)+5y=13$$

Multiply 4 times -\frac{4y}{5}+1.

$$-\frac{16}{5}y+4+5y=13$$

Add -\frac{16y}{5} to 5y.

$$\frac{9}{5}y+4=13$$

Subtract 4 from both sides of the equation.

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

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

$$y=5$$

Substitute 5 for y in x=-\frac{4}{5}y+1. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{4}{5}\times 5+1$$

Multiply -\frac{4}{5} times 5.

$$x=-4+1$$

Add 1 to -4.

$$x=-3$$

The system is now solved.

$$x=-3,y=5$$

Steps Using Matrices

Consider the first equation. Subtract 8x from both sides.

$$3x-4y-8x=-5$$

Combine 3x and -8x to get -5x.

$$-5x-4y=-5$$

Consider the second equation. Subtract 3y from both sides.

$$4x+8y-3y=13$$

Combine 8y and -3y to get 5y.

$$4x+5y=13$$

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

$$-5x-4y=-5,4x+5y=13$$

Write the equations in matrix form.

$$\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\\13\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-5&-4\\4&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right))\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right))\left(\begin{matrix}-5\\13\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right))\left(\begin{matrix}-5\\13\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-4\\4&5\end{matrix}\right))\left(\begin{matrix}-5\\13\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{-5\times 5-\left(-4\times 4\right)}&-\frac{-4}{-5\times 5-\left(-4\times 4\right)}\\-\frac{4}{-5\times 5-\left(-4\times 4\right)}&-\frac{5}{-5\times 5-\left(-4\times 4\right)}\end{matrix}\right)\left(\begin{matrix}-5\\13\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{9}&-\frac{4}{9}\\\frac{4}{9}&\frac{5}{9}\end{matrix}\right)\left(\begin{matrix}-5\\13\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{9}\left(-5\right)-\frac{4}{9}\times 13\\\frac{4}{9}\left(-5\right)+\frac{5}{9}\times 13\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-3,y=5$$

Steps Using Elimination

Consider the first equation. Subtract 8x from both sides.

$$3x-4y-8x=-5$$

Combine 3x and -8x to get -5x.

$$-5x-4y=-5$$

Consider the second equation. Subtract 3y from both sides.

$$4x+8y-3y=13$$

Combine 8y and -3y to get 5y.

$$4x+5y=13$$

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.

$$-5x-4y=-5,4x+5y=13$$

To make -5x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by -5.

$$4\left(-5\right)x+4\left(-4\right)y=4\left(-5\right),-5\times 4x-5\times 5y=-5\times 13$$

Simplify.

$$-20x-16y=-20,-20x-25y=-65$$

Subtract -20x-25y=-65 from -20x-16y=-20 by subtracting like terms on each side of the equal sign.

$$-20x+20x-16y+25y=-20+65$$

Add -20x to 20x. Terms -20x and 20x cancel out, leaving an equation with only one variable that can be solved.

$$-16y+25y=-20+65$$

Add -16y to 25y.

$$9y=-20+65$$

Add -20 to 65.

$$9y=45$$

Divide both sides by 9.

$$y=5$$

Substitute 5 for y in 4x+5y=13. Because the resulting equation contains only one variable, you can solve for x directly.

$$4x+5\times 5=13$$

Multiply 5 times 5.

$$4x+25=13$$

Subtract 25 from both sides of the equation.

$$4x=-12$$

Divide both sides by 4.

$$x=-3$$

The system is now solved.

$$x=-3,y=5$$