Solve {l}{-5x-10y=15}{7x+10y=-29}

$$\left. \begin{array} { l } { - 5 x - 10 y = 15 } \\ { 7 x + 10 y = - 29 } \end{array} \right.$$

Solve for x, y

$$x=-7$$

$$y=2$$

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.

$$-5x-10y=15,7x+10y=-29$$

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

$$-5x-10y=15$$

Add 10y to both sides of the equation.

$$-5x=10y+15$$

Divide both sides by -5.

$$x=-\frac{1}{5}\left(10y+15\right)$$

Multiply -\frac{1}{5} times 10y+15.

$$x=-2y-3$$

Substitute -2y-3 for x in the other equation, 7x+10y=-29.

$$7\left(-2y-3\right)+10y=-29$$

Multiply 7 times -2y-3.

$$-14y-21+10y=-29$$

Add -14y to 10y.

$$-4y-21=-29$$

Add 21 to both sides of the equation.

$$-4y=-8$$

Divide both sides by -4.

$$y=2$$

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

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

Multiply -2 times 2.

$$x=-4-3$$

Add -3 to -4.

$$x=-7$$

The system is now solved.

$$x=-7,y=2$$

Steps Using Matrices

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

$$-5x-10y=15,7x+10y=-29$$

Write the equations in matrix form.

$$\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}15\\-29\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-5&-10\\7&10\end{matrix}\right).

$$inverse(\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right))\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right))\left(\begin{matrix}15\\-29\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right))\left(\begin{matrix}15\\-29\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-5&-10\\7&10\end{matrix}\right))\left(\begin{matrix}15\\-29\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{-5\times 10-\left(-10\times 7\right)}&-\frac{-10}{-5\times 10-\left(-10\times 7\right)}\\-\frac{7}{-5\times 10-\left(-10\times 7\right)}&-\frac{5}{-5\times 10-\left(-10\times 7\right)}\end{matrix}\right)\left(\begin{matrix}15\\-29\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}&\frac{1}{2}\\-\frac{7}{20}&-\frac{1}{4}\end{matrix}\right)\left(\begin{matrix}15\\-29\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\times 15+\frac{1}{2}\left(-29\right)\\-\frac{7}{20}\times 15-\frac{1}{4}\left(-29\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-7\\2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-7,y=2$$

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.

$$-5x-10y=15,7x+10y=-29$$

To make -5x and 7x equal, multiply all terms on each side of the first equation by 7 and all terms on each side of the second by -5.

$$7\left(-5\right)x+7\left(-10\right)y=7\times 15,-5\times 7x-5\times 10y=-5\left(-29\right)$$

Simplify.

$$-35x-70y=105,-35x-50y=145$$

Subtract -35x-50y=145 from -35x-70y=105 by subtracting like terms on each side of the equal sign.

$$-35x+35x-70y+50y=105-145$$

Add -35x to 35x. Terms -35x and 35x cancel out, leaving an equation with only one variable that can be solved.

$$-70y+50y=105-145$$

Add -70y to 50y.

$$-20y=105-145$$

Add 105 to -145.

$$-20y=-40$$

Divide both sides by -20.

$$y=2$$

Substitute 2 for y in 7x+10y=-29. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x+10\times 2=-29$$

Multiply 10 times 2.

$$7x+20=-29$$

Subtract 20 from both sides of the equation.

$$7x=-49$$

Divide both sides by 7.

$$x=-7$$

The system is now solved.

$$x=-7,y=2$$