Solve {c}{x+y=14}{x/10+10y=x+y-18}

$$\left. \begin{array} { c } { x + y = 14 } \\ { \frac { x } { 10 } + 10 y = x + y - 18 } \end{array} \right.$$

Solve for x, y

$$x = \frac{160}{11} = 14\frac{6}{11} \approx 14.545454545$$

$$y=-\frac{6}{11}\approx -0.545454545$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 10.

$$x+100y=10x+10y-180$$

Subtract 10x from both sides.

$$x+100y-10x=10y-180$$

Combine x and -10x to get -9x.

$$-9x+100y=10y-180$$

Subtract 10y from both sides.

$$-9x+100y-10y=-180$$

Combine 100y and -10y to get 90y.

$$-9x+90y=-180$$

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.

$$x+y=14,-9x+90y=-180$$

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

$$x+y=14$$

Subtract y from both sides of the equation.

$$x=-y+14$$

Substitute -y+14 for x in the other equation, -9x+90y=-180.

$$-9\left(-y+14\right)+90y=-180$$

Multiply -9 times -y+14.

$$9y-126+90y=-180$$

Add 9y to 90y.

$$99y-126=-180$$

Add 126 to both sides of the equation.

$$99y=-54$$

Divide both sides by 99.

$$y=-\frac{6}{11}$$

Substitute -\frac{6}{11} for y in x=-y+14. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\left(-\frac{6}{11}\right)+14$$

Multiply -1 times -\frac{6}{11}.

$$x=\frac{6}{11}+14$$

Add 14 to \frac{6}{11}.

$$x=\frac{160}{11}$$

The system is now solved.

$$x=\frac{160}{11},y=-\frac{6}{11}$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 10.

$$x+100y=10x+10y-180$$

Subtract 10x from both sides.

$$x+100y-10x=10y-180$$

Combine x and -10x to get -9x.

$$-9x+100y=10y-180$$

Subtract 10y from both sides.

$$-9x+100y-10y=-180$$

Combine 100y and -10y to get 90y.

$$-9x+90y=-180$$

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

$$x+y=14,-9x+90y=-180$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\-9&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\-180\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-9&90\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\-9&90\end{matrix}\right))\left(\begin{matrix}1&1\\-9&90\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-9&90\end{matrix}\right))\left(\begin{matrix}14\\-180\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-9&90\end{matrix}\right))\left(\begin{matrix}14\\-180\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-9&90\end{matrix}\right))\left(\begin{matrix}14\\-180\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{90}{90-\left(-9\right)}&-\frac{1}{90-\left(-9\right)}\\-\frac{-9}{90-\left(-9\right)}&\frac{1}{90-\left(-9\right)}\end{matrix}\right)\left(\begin{matrix}14\\-180\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{11}&-\frac{1}{99}\\\frac{1}{11}&\frac{1}{99}\end{matrix}\right)\left(\begin{matrix}14\\-180\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{11}\times 14-\frac{1}{99}\left(-180\right)\\\frac{1}{11}\times 14+\frac{1}{99}\left(-180\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{160}{11}\\-\frac{6}{11}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{160}{11},y=-\frac{6}{11}$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 10.

$$x+100y=10x+10y-180$$

Subtract 10x from both sides.

$$x+100y-10x=10y-180$$

Combine x and -10x to get -9x.

$$-9x+100y=10y-180$$

Subtract 10y from both sides.

$$-9x+100y-10y=-180$$

Combine 100y and -10y to get 90y.

$$-9x+90y=-180$$

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.

$$x+y=14,-9x+90y=-180$$

To make x and -9x equal, multiply all terms on each side of the first equation by -9 and all terms on each side of the second by 1.

$$-9x-9y=-9\times 14,-9x+90y=-180$$

Simplify.

$$-9x-9y=-126,-9x+90y=-180$$

Subtract -9x+90y=-180 from -9x-9y=-126 by subtracting like terms on each side of the equal sign.

$$-9x+9x-9y-90y=-126+180$$

Add -9x to 9x. Terms -9x and 9x cancel out, leaving an equation with only one variable that can be solved.

$$-9y-90y=-126+180$$

Add -9y to -90y.

$$-99y=-126+180$$

Add -126 to 180.

$$-99y=54$$

Divide both sides by -99.

$$y=-\frac{6}{11}$$

Substitute -\frac{6}{11} for y in -9x+90y=-180. Because the resulting equation contains only one variable, you can solve for x directly.

$$-9x+90\left(-\frac{6}{11}\right)=-180$$

Multiply 90 times -\frac{6}{11}.

$$-9x-\frac{540}{11}=-180$$

Add \frac{540}{11} to both sides of the equation.

$$-9x=-\frac{1440}{11}$$

Divide both sides by -9.

$$x=\frac{160}{11}$$

The system is now solved.

$$x=\frac{160}{11},y=-\frac{6}{11}$$