Solve {r}{6x+120+5(y-20)=216}{x+y=40}

$$\left. \begin{array} { r } { 6 x + 120 + 5 ( y - 20 ) = 216 } \\ { x + y = 40 } \end{array} \right.$$

Solve for x, y

$$x=-4$$

$$y=44$$

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.

$$6x+5\left(y-20\right)+120=216,x+y=40$$

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

$$6x+5\left(y-20\right)+120=216$$

Multiply 5 times y-20.

$$6x+5y-100+120=216$$

Add -100 to 120.

$$6x+5y+20=216$$

Subtract 20 from both sides of the equation.

$$6x+5y=196$$

Subtract 5y from both sides of the equation.

$$6x=-5y+196$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-5y+196\right)$$

Multiply \frac{1}{6} times -5y+196.

$$x=-\frac{5}{6}y+\frac{98}{3}$$

Substitute -\frac{5y}{6}+\frac{98}{3} for x in the other equation, x+y=40.

$$-\frac{5}{6}y+\frac{98}{3}+y=40$$

Add -\frac{5y}{6} to y.

$$\frac{1}{6}y+\frac{98}{3}=40$$

Subtract \frac{98}{3} from both sides of the equation.

$$\frac{1}{6}y=\frac{22}{3}$$

Multiply both sides by 6.

$$y=44$$

Substitute 44 for y in x=-\frac{5}{6}y+\frac{98}{3}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{6}\times 44+\frac{98}{3}$$

Multiply -\frac{5}{6} times 44.

$$x=\frac{-110+98}{3}$$

Add \frac{98}{3} to -\frac{110}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-4$$

The system is now solved.

$$x=-4,y=44$$

Steps Using Matrices

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

$$6x+5\left(y-20\right)+120=216,x+y=40$$

Simplify the first equation to put it in standard form.

$$6x+5\left(y-20\right)+120=216$$

Multiply 5 times y-20.

$$6x+5y-100+120=216$$

Add -100 to 120.

$$6x+5y+20=216$$

Subtract 20 from both sides of the equation.

$$6x+5y=196$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}196\\40\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&5\\1&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&5\\1&1\end{matrix}\right))\left(\begin{matrix}6&5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\1&1\end{matrix}\right))\left(\begin{matrix}196\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\1&1\end{matrix}\right))\left(\begin{matrix}196\\40\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&5\\1&1\end{matrix}\right))\left(\begin{matrix}196\\40\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6-5}&-\frac{5}{6-5}\\-\frac{1}{6-5}&\frac{6}{6-5}\end{matrix}\right)\left(\begin{matrix}196\\40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}1&-5\\-1&6\end{matrix}\right)\left(\begin{matrix}196\\40\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}196-5\times 40\\-196+6\times 40\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\\44\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-4,y=44$$