Solve {l}{40-2x-y=2x+5y+2}{8x-3y-4=40-2x-y}

$$\left\{ \begin{array} { l } { 40 - 2 x - y = 2 x + 5 y + 2 } \\ { 8 x - 3 y - 4 = 40 - 2 x - y } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=3$$

Steps Using Substitution

Consider the first equation. Subtract 2x from both sides.

$$40-2x-y-2x=5y+2$$

Combine -2x and -2x to get -4x.

$$40-4x-y=5y+2$$

Subtract 5y from both sides.

$$40-4x-y-5y=2$$

Combine -y and -5y to get -6y.

$$40-4x-6y=2$$

Subtract 40 from both sides.

$$-4x-6y=2-40$$

Subtract 40 from 2 to get -38.

$$-4x-6y=-38$$

Consider the second equation. Add 2x to both sides.

$$8x-3y-4+2x=40-y$$

Combine 8x and 2x to get 10x.

$$10x-3y-4=40-y$$

Add y to both sides.

$$10x-3y-4+y=40$$

Combine -3y and y to get -2y.

$$10x-2y-4=40$$

Add 4 to both sides.

$$10x-2y=40+4$$

Add 40 and 4 to get 44.

$$10x-2y=44$$

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.

$$-4x-6y=-38,10x-2y=44$$

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

$$-4x-6y=-38$$

Add 6y to both sides of the equation.

$$-4x=6y-38$$

Divide both sides by -4.

$$x=-\frac{1}{4}\left(6y-38\right)$$

Multiply -\frac{1}{4} times 6y-38.

$$x=-\frac{3}{2}y+\frac{19}{2}$$

Substitute \frac{-3y+19}{2} for x in the other equation, 10x-2y=44.

$$10\left(-\frac{3}{2}y+\frac{19}{2}\right)-2y=44$$

Multiply 10 times \frac{-3y+19}{2}.

$$-15y+95-2y=44$$

Add -15y to -2y.

$$-17y+95=44$$

Subtract 95 from both sides of the equation.

$$-17y=-51$$

Divide both sides by -17.

$$y=3$$

Substitute 3 for y in x=-\frac{3}{2}y+\frac{19}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{3}{2}\times 3+\frac{19}{2}$$

Multiply -\frac{3}{2} times 3.

$$x=\frac{-9+19}{2}$$

Add \frac{19}{2} to -\frac{9}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=3$$

Steps Using Matrices

Consider the first equation. Subtract 2x from both sides.

$$40-2x-y-2x=5y+2$$

Combine -2x and -2x to get -4x.

$$40-4x-y=5y+2$$

Subtract 5y from both sides.

$$40-4x-y-5y=2$$

Combine -y and -5y to get -6y.

$$40-4x-6y=2$$

Subtract 40 from both sides.

$$-4x-6y=2-40$$

Subtract 40 from 2 to get -38.

$$-4x-6y=-38$$

Consider the second equation. Add 2x to both sides.

$$8x-3y-4+2x=40-y$$

Combine 8x and 2x to get 10x.

$$10x-3y-4=40-y$$

Add y to both sides.

$$10x-3y-4+y=40$$

Combine -3y and y to get -2y.

$$10x-2y-4=40$$

Add 4 to both sides.

$$10x-2y=40+4$$

Add 40 and 4 to get 44.

$$10x-2y=44$$

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

$$-4x-6y=-38,10x-2y=44$$

Write the equations in matrix form.

$$\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-38\\44\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right).

$$inverse(\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right))\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right))\left(\begin{matrix}-38\\44\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right))\left(\begin{matrix}-38\\44\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-4&-6\\10&-2\end{matrix}\right))\left(\begin{matrix}-38\\44\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{-4\left(-2\right)-\left(-6\times 10\right)}&-\frac{-6}{-4\left(-2\right)-\left(-6\times 10\right)}\\-\frac{10}{-4\left(-2\right)-\left(-6\times 10\right)}&-\frac{4}{-4\left(-2\right)-\left(-6\times 10\right)}\end{matrix}\right)\left(\begin{matrix}-38\\44\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{34}&\frac{3}{34}\\-\frac{5}{34}&-\frac{1}{17}\end{matrix}\right)\left(\begin{matrix}-38\\44\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{34}\left(-38\right)+\frac{3}{34}\times 44\\-\frac{5}{34}\left(-38\right)-\frac{1}{17}\times 44\end{matrix}\right)$$

Do the arithmetic.

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

Extract the matrix elements x and y.

$$x=5,y=3$$

Steps Using Elimination

Consider the first equation. Subtract 2x from both sides.

$$40-2x-y-2x=5y+2$$

Combine -2x and -2x to get -4x.

$$40-4x-y=5y+2$$

Subtract 5y from both sides.

$$40-4x-y-5y=2$$

Combine -y and -5y to get -6y.

$$40-4x-6y=2$$

Subtract 40 from both sides.

$$-4x-6y=2-40$$

Subtract 40 from 2 to get -38.

$$-4x-6y=-38$$

Consider the second equation. Add 2x to both sides.

$$8x-3y-4+2x=40-y$$

Combine 8x and 2x to get 10x.

$$10x-3y-4=40-y$$

Add y to both sides.

$$10x-3y-4+y=40$$

Combine -3y and y to get -2y.

$$10x-2y-4=40$$

Add 4 to both sides.

$$10x-2y=40+4$$

Add 40 and 4 to get 44.

$$10x-2y=44$$

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.

$$-4x-6y=-38,10x-2y=44$$

To make -4x 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 -4.

$$10\left(-4\right)x+10\left(-6\right)y=10\left(-38\right),-4\times 10x-4\left(-2\right)y=-4\times 44$$

Simplify.

$$-40x-60y=-380,-40x+8y=-176$$

Subtract -40x+8y=-176 from -40x-60y=-380 by subtracting like terms on each side of the equal sign.

$$-40x+40x-60y-8y=-380+176$$

Add -40x to 40x. Terms -40x and 40x cancel out, leaving an equation with only one variable that can be solved.

$$-60y-8y=-380+176$$

Add -60y to -8y.

$$-68y=-380+176$$

Add -380 to 176.

$$-68y=-204$$

Divide both sides by -68.

$$y=3$$

Substitute 3 for y in 10x-2y=44. Because the resulting equation contains only one variable, you can solve for x directly.

$$10x-2\times 3=44$$

Multiply -2 times 3.

$$10x-6=44$$

Add 6 to both sides of the equation.

$$10x=50$$

Divide both sides by 10.

$$x=5$$

The system is now solved.

$$x=5,y=3$$