Solve {l}{x+y=40}{30x+25y=1125}

$$\left. \begin{array} { l } { x + y = 40 } \\ { 30 x + 25 y = 1125 } \end{array} \right.$$

Solve for x, y

$$x=25$$

$$y=15$$

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.

$$x+y=40,30x+25y=1125$$

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

$$x+y=40$$

Subtract y from both sides of the equation.

$$x=-y+40$$

Substitute -y+40 for x in the other equation, 30x+25y=1125.

$$30\left(-y+40\right)+25y=1125$$

Multiply 30 times -y+40.

$$-30y+1200+25y=1125$$

Add -30y to 25y.

$$-5y+1200=1125$$

Subtract 1200 from both sides of the equation.

$$-5y=-75$$

Divide both sides by -5.

$$y=15$$

Substitute 15 for y in x=-y+40. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-15+40$$

Add 40 to -15.

$$x=25$$

The system is now solved.

$$x=25,y=15$$

Steps Using Matrices

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

$$x+y=40,30x+25y=1125$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\30&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}40\\1125\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\30&25\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\30&25\end{matrix}\right))\left(\begin{matrix}1&1\\30&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&25\end{matrix}\right))\left(\begin{matrix}40\\1125\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&25\end{matrix}\right))\left(\begin{matrix}40\\1125\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\30&25\end{matrix}\right))\left(\begin{matrix}40\\1125\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{25-30}&-\frac{1}{25-30}\\-\frac{30}{25-30}&\frac{1}{25-30}\end{matrix}\right)\left(\begin{matrix}40\\1125\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5&\frac{1}{5}\\6&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}40\\1125\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-5\times 40+\frac{1}{5}\times 1125\\6\times 40-\frac{1}{5}\times 1125\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}25\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=25,y=15$$

Steps Using Elimination

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.

$$x+y=40,30x+25y=1125$$

To make x and 30x equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 1.

$$30x+30y=30\times 40,30x+25y=1125$$

Simplify.

$$30x+30y=1200,30x+25y=1125$$

Subtract 30x+25y=1125 from 30x+30y=1200 by subtracting like terms on each side of the equal sign.

$$30x-30x+30y-25y=1200-1125$$

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

$$30y-25y=1200-1125$$

Add 30y to -25y.

$$5y=1200-1125$$

Add 1200 to -1125.

$$5y=75$$

Divide both sides by 5.

$$y=15$$

Substitute 15 for y in 30x+25y=1125. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x+25\times 15=1125$$

Multiply 25 times 15.

$$30x+375=1125$$

Subtract 375 from both sides of the equation.

$$30x=750$$

Divide both sides by 30.

$$x=25$$

The system is now solved.

$$x=25,y=15$$