Solve {l}{10x+25y=600}{15x+30y=750}

$$\left\{ \begin{array} { l } { 10 x + 25 y = 600 } \\ { 15 x + 30 y = 750 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=20$$

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.

$$10x+25y=600,15x+30y=750$$

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

$$10x+25y=600$$

Subtract 25y from both sides of the equation.

$$10x=-25y+600$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-25y+600\right)$$

Multiply \frac{1}{10} times -25y+600.

$$x=-\frac{5}{2}y+60$$

Substitute -\frac{5y}{2}+60 for x in the other equation, 15x+30y=750.

$$15\left(-\frac{5}{2}y+60\right)+30y=750$$

Multiply 15 times -\frac{5y}{2}+60.

$$-\frac{75}{2}y+900+30y=750$$

Add -\frac{75y}{2} to 30y.

$$-\frac{15}{2}y+900=750$$

Subtract 900 from both sides of the equation.

$$-\frac{15}{2}y=-150$$

Divide both sides of the equation by -\frac{15}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=20$$

Substitute 20 for y in x=-\frac{5}{2}y+60. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{5}{2}\times 20+60$$

Multiply -\frac{5}{2} times 20.

$$x=-50+60$$

Add 60 to -50.

$$x=10$$

The system is now solved.

$$x=10,y=20$$

Steps Using Matrices

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

$$10x+25y=600,15x+30y=750$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&25\\15&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}600\\750\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&25\\15&30\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&25\\15&30\end{matrix}\right))\left(\begin{matrix}10&25\\15&30\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&25\\15&30\end{matrix}\right))\left(\begin{matrix}600\\750\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&25\\15&30\end{matrix}\right))\left(\begin{matrix}600\\750\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&25\\15&30\end{matrix}\right))\left(\begin{matrix}600\\750\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{30}{10\times 30-25\times 15}&-\frac{25}{10\times 30-25\times 15}\\-\frac{15}{10\times 30-25\times 15}&\frac{10}{10\times 30-25\times 15}\end{matrix}\right)\left(\begin{matrix}600\\750\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5}&\frac{1}{3}\\\frac{1}{5}&-\frac{2}{15}\end{matrix}\right)\left(\begin{matrix}600\\750\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{5}\times 600+\frac{1}{3}\times 750\\\frac{1}{5}\times 600-\frac{2}{15}\times 750\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\20\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=20$$

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.

$$10x+25y=600,15x+30y=750$$

To make 10x and 15x equal, multiply all terms on each side of the first equation by 15 and all terms on each side of the second by 10.

$$15\times 10x+15\times 25y=15\times 600,10\times 15x+10\times 30y=10\times 750$$

Simplify.

$$150x+375y=9000,150x+300y=7500$$

Subtract 150x+300y=7500 from 150x+375y=9000 by subtracting like terms on each side of the equal sign.

$$150x-150x+375y-300y=9000-7500$$

Add 150x to -150x. Terms 150x and -150x cancel out, leaving an equation with only one variable that can be solved.

$$375y-300y=9000-7500$$

Add 375y to -300y.

$$75y=9000-7500$$

Add 9000 to -7500.

$$75y=1500$$

Divide both sides by 75.

$$y=20$$

Substitute 20 for y in 15x+30y=750. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x+30\times 20=750$$

Multiply 30 times 20.

$$15x+600=750$$

Subtract 600 from both sides of the equation.

$$15x=150$$

Divide both sides by 15.

$$x=10$$

The system is now solved.

$$x=10,y=20$$