Solve {l}{3x+2y=60}{5x+3y=95}

$$\left\{ \begin{array} { l } { 3 x + 2 y = 60 } \\ { 5 x + 3 y = 95 } \end{array} \right.$$

Solve for x, y

$$x=10$$

$$y=15$$

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.

$$3x+2y=60,5x+3y=95$$

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

$$3x+2y=60$$

Subtract 2y from both sides of the equation.

$$3x=-2y+60$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(-2y+60\right)$$

Multiply \frac{1}{3} times -2y+60.

$$x=-\frac{2}{3}y+20$$

Substitute -\frac{2y}{3}+20 for x in the other equation, 5x+3y=95.

$$5\left(-\frac{2}{3}y+20\right)+3y=95$$

Multiply 5 times -\frac{2y}{3}+20.

$$-\frac{10}{3}y+100+3y=95$$

Add -\frac{10y}{3} to 3y.

$$-\frac{1}{3}y+100=95$$

Subtract 100 from both sides of the equation.

$$-\frac{1}{3}y=-5$$

Multiply both sides by -3.

$$y=15$$

Substitute 15 for y in x=-\frac{2}{3}y+20. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{2}{3}\times 15+20$$

Multiply -\frac{2}{3} times 15.

$$x=-10+20$$

Add 20 to -10.

$$x=10$$

The system is now solved.

$$x=10,y=15$$

Steps Using Matrices

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

$$3x+2y=60,5x+3y=95$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&2\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\95\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&2\\5&3\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}3&2\\5&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}60\\95\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}60\\95\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&2\\5&3\end{matrix}\right))\left(\begin{matrix}60\\95\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{3\times 3-2\times 5}&-\frac{2}{3\times 3-2\times 5}\\-\frac{5}{3\times 3-2\times 5}&\frac{3}{3\times 3-2\times 5}\end{matrix}\right)\left(\begin{matrix}60\\95\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3&2\\5&-3\end{matrix}\right)\left(\begin{matrix}60\\95\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 60+2\times 95\\5\times 60-3\times 95\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}10\\15\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=10,y=15$$

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.

$$3x+2y=60,5x+3y=95$$

To make 3x and 5x equal, multiply all terms on each side of the first equation by 5 and all terms on each side of the second by 3.

$$5\times 3x+5\times 2y=5\times 60,3\times 5x+3\times 3y=3\times 95$$

Simplify.

$$15x+10y=300,15x+9y=285$$

Subtract 15x+9y=285 from 15x+10y=300 by subtracting like terms on each side of the equal sign.

$$15x-15x+10y-9y=300-285$$

Add 15x to -15x. Terms 15x and -15x cancel out, leaving an equation with only one variable that can be solved.

$$10y-9y=300-285$$

Add 10y to -9y.

$$y=300-285$$

Add 300 to -285.

$$y=15$$

Substitute 15 for y in 5x+3y=95. Because the resulting equation contains only one variable, you can solve for x directly.

$$5x+3\times 15=95$$

Multiply 3 times 15.

$$5x+45=95$$

Subtract 45 from both sides of the equation.

$$5x=50$$

Divide both sides by 5.

$$x=10$$

The system is now solved.

$$x=10,y=15$$