Solve {l}{3x-6y=-15}{15x+25y=89}

$$\left\{ \begin{array} { l } { 3 x - 6 y = - 15 } \\ { 15 x + 25 y = 89 } \end{array} \right.$$

Solve for x, y

$$x=\frac{53}{55}\approx 0.963636364$$

$$y = \frac{164}{55} = 2\frac{54}{55} \approx 2.981818182$$

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.

$$3x-6y=-15,15x+25y=89$$

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

$$3x-6y=-15$$

Add 6y to both sides of the equation.

$$3x=6y-15$$

Divide both sides by 3.

$$x=\frac{1}{3}\left(6y-15\right)$$

Multiply \frac{1}{3} times 6y-15.

$$x=2y-5$$

Substitute 2y-5 for x in the other equation, 15x+25y=89.

$$15\left(2y-5\right)+25y=89$$

Multiply 15 times 2y-5.

$$30y-75+25y=89$$

Add 30y to 25y.

$$55y-75=89$$

Add 75 to both sides of the equation.

$$55y=164$$

Divide both sides by 55.

$$y=\frac{164}{55}$$

Substitute \frac{164}{55} for y in x=2y-5. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2\times \frac{164}{55}-5$$

Multiply 2 times \frac{164}{55}.

$$x=\frac{328}{55}-5$$

Add -5 to \frac{328}{55}.

$$x=\frac{53}{55}$$

The system is now solved.

$$x=\frac{53}{55},y=\frac{164}{55}$$

Steps Using Matrices

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

$$3x-6y=-15,15x+25y=89$$

Write the equations in matrix form.

$$\left(\begin{matrix}3&-6\\15&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-15\\89\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&-6\\15&25\end{matrix}\right).

$$inverse(\left(\begin{matrix}3&-6\\15&25\end{matrix}\right))\left(\begin{matrix}3&-6\\15&25\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-6\\15&25\end{matrix}\right))\left(\begin{matrix}-15\\89\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-6\\15&25\end{matrix}\right))\left(\begin{matrix}-15\\89\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&-6\\15&25\end{matrix}\right))\left(\begin{matrix}-15\\89\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{3\times 25-\left(-6\times 15\right)}&-\frac{-6}{3\times 25-\left(-6\times 15\right)}\\-\frac{15}{3\times 25-\left(-6\times 15\right)}&\frac{3}{3\times 25-\left(-6\times 15\right)}\end{matrix}\right)\left(\begin{matrix}-15\\89\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{33}&\frac{2}{55}\\-\frac{1}{11}&\frac{1}{55}\end{matrix}\right)\left(\begin{matrix}-15\\89\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{33}\left(-15\right)+\frac{2}{55}\times 89\\-\frac{1}{11}\left(-15\right)+\frac{1}{55}\times 89\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{53}{55}\\\frac{164}{55}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{53}{55},y=\frac{164}{55}$$

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.

$$3x-6y=-15,15x+25y=89$$

To make 3x 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 3.

$$15\times 3x+15\left(-6\right)y=15\left(-15\right),3\times 15x+3\times 25y=3\times 89$$

Simplify.

$$45x-90y=-225,45x+75y=267$$

Subtract 45x+75y=267 from 45x-90y=-225 by subtracting like terms on each side of the equal sign.

$$45x-45x-90y-75y=-225-267$$

Add 45x to -45x. Terms 45x and -45x cancel out, leaving an equation with only one variable that can be solved.

$$-90y-75y=-225-267$$

Add -90y to -75y.

$$-165y=-225-267$$

Add -225 to -267.

$$-165y=-492$$

Divide both sides by -165.

$$y=\frac{164}{55}$$

Substitute \frac{164}{55} for y in 15x+25y=89. Because the resulting equation contains only one variable, you can solve for x directly.

$$15x+25\times \frac{164}{55}=89$$

Multiply 25 times \frac{164}{55}.

$$15x+\frac{820}{11}=89$$

Subtract \frac{820}{11} from both sides of the equation.

$$15x=\frac{159}{11}$$

Divide both sides by 15.

$$x=\frac{53}{55}$$

The system is now solved.

$$x=\frac{53}{55},y=\frac{164}{55}$$