Solve {l}{x+y=20}{x+5=11/2y}

$$\left. \begin{array} { l } { x + y = 20 } \\ { x + 5 = 1 \frac { 1 } { 2 } y } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=10$$

Steps Using Substitution

Consider the second equation. Multiply both sides of the equation by 2.

$$2x+10=\left(1\times 2+1\right)y$$

Multiply 1 and 2 to get 2.

$$2x+10=\left(2+1\right)y$$

Add 2 and 1 to get 3.

$$2x+10=3y$$

Subtract 3y from both sides.

$$2x+10-3y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$2x-3y=-10$$

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.

$$x+y=20,2x-3y=-10$$

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

$$x+y=20$$

Subtract y from both sides of the equation.

$$x=-y+20$$

Substitute -y+20 for x in the other equation, 2x-3y=-10.

$$2\left(-y+20\right)-3y=-10$$

Multiply 2 times -y+20.

$$-2y+40-3y=-10$$

Add -2y to -3y.

$$-5y+40=-10$$

Subtract 40 from both sides of the equation.

$$-5y=-50$$

Divide both sides by -5.

$$y=10$$

Substitute 10 for y in x=-y+20. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-10+20$$

Add 20 to -10.

$$x=10$$

The system is now solved.

$$x=10,y=10$$

Steps Using Matrices

Consider the second equation. Multiply both sides of the equation by 2.

$$2x+10=\left(1\times 2+1\right)y$$

Multiply 1 and 2 to get 2.

$$2x+10=\left(2+1\right)y$$

Add 2 and 1 to get 3.

$$2x+10=3y$$

Subtract 3y from both sides.

$$2x+10-3y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$2x-3y=-10$$

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

$$x+y=20,2x-3y=-10$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\-10\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2&-3\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}1&1\\2&-3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}20\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}20\\-10\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2&-3\end{matrix}\right))\left(\begin{matrix}20\\-10\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{3}{-3-2}&-\frac{1}{-3-2}\\-\frac{2}{-3-2}&\frac{1}{-3-2}\end{matrix}\right)\left(\begin{matrix}20\\-10\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}&\frac{1}{5}\\\frac{2}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}20\\-10\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{5}\times 20+\frac{1}{5}\left(-10\right)\\\frac{2}{5}\times 20-\frac{1}{5}\left(-10\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\10\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=10$$

Steps Using Elimination

Consider the second equation. Multiply both sides of the equation by 2.

$$2x+10=\left(1\times 2+1\right)y$$

Multiply 1 and 2 to get 2.

$$2x+10=\left(2+1\right)y$$

Add 2 and 1 to get 3.

$$2x+10=3y$$

Subtract 3y from both sides.

$$2x+10-3y=0$$

Subtract 10 from both sides. Anything subtracted from zero gives its negation.

$$2x-3y=-10$$

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.

$$x+y=20,2x-3y=-10$$

To make x and 2x equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 1.

$$2x+2y=2\times 20,2x-3y=-10$$

Simplify.

$$2x+2y=40,2x-3y=-10$$

Subtract 2x-3y=-10 from 2x+2y=40 by subtracting like terms on each side of the equal sign.

$$2x-2x+2y+3y=40+10$$

Add 2x to -2x. Terms 2x and -2x cancel out, leaving an equation with only one variable that can be solved.

$$2y+3y=40+10$$

Add 2y to 3y.

$$5y=40+10$$

Add 40 to 10.

$$5y=50$$

Divide both sides by 5.

$$y=10$$

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

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

Multiply -3 times 10.

$$2x-30=-10$$

Add 30 to both sides of the equation.

$$2x=20$$

Divide both sides by 2.

$$x=10$$

The system is now solved.

$$x=10,y=10$$