Solve {l}{2x-96=4y-x}{3y+10x=y+90}

$$\left. \begin{array} { l } { 2 x - 96 = 4 y - x } \\ { 3 y + 10 x = y + 90 } \end{array} \right.$$

Solve for x, y

$$x=12$$

$$y=-15$$

Steps Using Substitution

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

$$2x-96-4y=-x$$

Add x to both sides.

$$2x-96-4y+x=0$$

Combine 2x and x to get 3x.

$$3x-96-4y=0$$

Add 96 to both sides. Anything plus zero gives itself.

$$3x-4y=96$$

Consider the second equation. Subtract y from both sides.

$$3y+10x-y=90$$

Combine 3y and -y to get 2y.

$$2y+10x=90$$

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.

$$3x-4y=96,10x+2y=90$$

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

$$3x-4y=96$$

Add 4y to both sides of the equation.

$$3x=4y+96$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(4y+96\right)$$

Multiply \frac{1}{3} times 96+4y.

$$x=\frac{4}{3}y+32$$

Substitute 32+\frac{4y}{3} for x in the other equation, 10x+2y=90.

$$10\left(\frac{4}{3}y+32\right)+2y=90$$

Multiply 10 times 32+\frac{4y}{3}.

$$\frac{40}{3}y+320+2y=90$$

Add \frac{40y}{3} to 2y.

$$\frac{46}{3}y+320=90$$

Subtract 320 from both sides of the equation.

$$\frac{46}{3}y=-230$$

Divide both sides of the equation by \frac{46}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-15$$

Substitute -15 for y in x=\frac{4}{3}y+32. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{4}{3}\left(-15\right)+32$$

Multiply \frac{4}{3} times -15.

$$x=-20+32$$

Add 32 to -20.

$$x=12$$

The system is now solved.

$$x=12,y=-15$$

Steps Using Matrices

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

$$2x-96-4y=-x$$

Add x to both sides.

$$2x-96-4y+x=0$$

Combine 2x and x to get 3x.

$$3x-96-4y=0$$

Add 96 to both sides. Anything plus zero gives itself.

$$3x-4y=96$$

Consider the second equation. Subtract y from both sides.

$$3y+10x-y=90$$

Combine 3y and -y to get 2y.

$$2y+10x=90$$

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

$$3x-4y=96,10x+2y=90$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-4\\10&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}96\\90\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-4\\10&2\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-4\\10&2\end{matrix}\right))\left(\begin{matrix}3&-4\\10&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\10&2\end{matrix}\right))\left(\begin{matrix}96\\90\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\10&2\end{matrix}\right))\left(\begin{matrix}96\\90\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-4\\10&2\end{matrix}\right))\left(\begin{matrix}96\\90\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3\times 2-\left(-4\times 10\right)}&-\frac{-4}{3\times 2-\left(-4\times 10\right)}\\-\frac{10}{3\times 2-\left(-4\times 10\right)}&\frac{3}{3\times 2-\left(-4\times 10\right)}\end{matrix}\right)\left(\begin{matrix}96\\90\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{23}&\frac{2}{23}\\-\frac{5}{23}&\frac{3}{46}\end{matrix}\right)\left(\begin{matrix}96\\90\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{23}\times 96+\frac{2}{23}\times 90\\-\frac{5}{23}\times 96+\frac{3}{46}\times 90\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12\\-15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=12,y=-15$$

Steps Using Elimination

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

$$2x-96-4y=-x$$

Add x to both sides.

$$2x-96-4y+x=0$$

Combine 2x and x to get 3x.

$$3x-96-4y=0$$

Add 96 to both sides. Anything plus zero gives itself.

$$3x-4y=96$$

Consider the second equation. Subtract y from both sides.

$$3y+10x-y=90$$

Combine 3y and -y to get 2y.

$$2y+10x=90$$

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.

$$3x-4y=96,10x+2y=90$$

To make 3x and 10x equal, multiply all terms on each side of the first equation by 10 and all terms on each side of the second by 3.

$$10\times 3x+10\left(-4\right)y=10\times 96,3\times 10x+3\times 2y=3\times 90$$

Simplify.

$$30x-40y=960,30x+6y=270$$

Subtract 30x+6y=270 from 30x-40y=960 by subtracting like terms on each side of the equal sign.

$$30x-30x-40y-6y=960-270$$

Add 30x to -30x. Terms 30x and -30x cancel out, leaving an equation with only one variable that can be solved.

$$-40y-6y=960-270$$

Add -40y to -6y.

$$-46y=960-270$$

Add 960 to -270.

$$-46y=690$$

Divide both sides by -46.

$$y=-15$$

Substitute -15 for y in 10x+2y=90. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x+2\left(-15\right)=90$$

Multiply 2 times -15.

$$10x-30=90$$

Add 30 to both sides of the equation.

$$10x=120$$

Divide both sides by 10.

$$x=12$$

The system is now solved.

$$x=12,y=-15$$