Solve {l}{12(x+2y)-8(2x+y)=2(5x-6y)}{20(x-4y)=-10}

$$\left\{ \begin{array} { l } { 12 ( x + 2 y ) - 8 ( 2 x + y ) = 2 ( 5 x - 6 y ) } \\ { 20 ( x - 4 y ) = - 10 } \end{array} \right.$$

Solve for x, y

$$x=\frac{1}{2}=0.5$$

$$y=\frac{1}{4}=0.25$$

Steps Using Substitution

Consider the first equation. Use the distributive property to multiply 12 by x+2y.

$$12x+24y-8\left(2x+y\right)=2\left(5x-6y\right)$$

Use the distributive property to multiply -8 by 2x+y.

$$12x+24y-16x-8y=2\left(5x-6y\right)$$

Combine 12x and -16x to get -4x.

$$-4x+24y-8y=2\left(5x-6y\right)$$

Combine 24y and -8y to get 16y.

$$-4x+16y=2\left(5x-6y\right)$$

Use the distributive property to multiply 2 by 5x-6y.

$$-4x+16y=10x-12y$$

Subtract 10x from both sides.

$$-4x+16y-10x=-12y$$

Combine -4x and -10x to get -14x.

$$-14x+16y=-12y$$

Add 12y to both sides.

$$-14x+16y+12y=0$$

Combine 16y and 12y to get 28y.

$$-14x+28y=0$$

Consider the second equation. Divide both sides by 20.

$$x-4y=\frac{-10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$x-4y=-\frac{1}{2}$$

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.

$$-14x+28y=0,x-4y=-\frac{1}{2}$$

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

$$-14x+28y=0$$

Subtract 28y from both sides of the equation.

$$-14x=-28y$$

Divide both sides by -14.

$$x=-\frac{1}{14}\left(-28\right)y$$

Multiply -\frac{1}{14} times -28y.

$$x=2y$$

Substitute 2y for x in the other equation, x-4y=-\frac{1}{2}.

$$2y-4y=-\frac{1}{2}$$

Add 2y to -4y.

$$-2y=-\frac{1}{2}$$

Divide both sides by -2.

$$y=\frac{1}{4}$$

Substitute \frac{1}{4} for y in x=2y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=2\times \frac{1}{4}$$

Multiply 2 times \frac{1}{4}.

$$x=\frac{1}{2}$$

The system is now solved.

$$x=\frac{1}{2},y=\frac{1}{4}$$

Steps Using Matrices

Consider the first equation. Use the distributive property to multiply 12 by x+2y.

$$12x+24y-8\left(2x+y\right)=2\left(5x-6y\right)$$

Use the distributive property to multiply -8 by 2x+y.

$$12x+24y-16x-8y=2\left(5x-6y\right)$$

Combine 12x and -16x to get -4x.

$$-4x+24y-8y=2\left(5x-6y\right)$$

Combine 24y and -8y to get 16y.

$$-4x+16y=2\left(5x-6y\right)$$

Use the distributive property to multiply 2 by 5x-6y.

$$-4x+16y=10x-12y$$

Subtract 10x from both sides.

$$-4x+16y-10x=-12y$$

Combine -4x and -10x to get -14x.

$$-14x+16y=-12y$$

Add 12y to both sides.

$$-14x+16y+12y=0$$

Combine 16y and 12y to get 28y.

$$-14x+28y=0$$

Consider the second equation. Divide both sides by 20.

$$x-4y=\frac{-10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$x-4y=-\frac{1}{2}$$

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

$$-14x+28y=0,x-4y=-\frac{1}{2}$$

Write the equations in matrix form.

$$\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}-14&28\\1&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right))\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right))\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right))\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}-14&28\\1&-4\end{matrix}\right))\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{-14\left(-4\right)-28}&-\frac{28}{-14\left(-4\right)-28}\\-\frac{1}{-14\left(-4\right)-28}&-\frac{14}{-14\left(-4\right)-28}\end{matrix}\right)\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{7}&-1\\-\frac{1}{28}&-\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}0\\-\frac{1}{2}\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\left(-\frac{1}{2}\right)\\-\frac{1}{2}\left(-\frac{1}{2}\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{2}\\\frac{1}{4}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{1}{2},y=\frac{1}{4}$$

Steps Using Elimination

Consider the first equation. Use the distributive property to multiply 12 by x+2y.

$$12x+24y-8\left(2x+y\right)=2\left(5x-6y\right)$$

Use the distributive property to multiply -8 by 2x+y.

$$12x+24y-16x-8y=2\left(5x-6y\right)$$

Combine 12x and -16x to get -4x.

$$-4x+24y-8y=2\left(5x-6y\right)$$

Combine 24y and -8y to get 16y.

$$-4x+16y=2\left(5x-6y\right)$$

Use the distributive property to multiply 2 by 5x-6y.

$$-4x+16y=10x-12y$$

Subtract 10x from both sides.

$$-4x+16y-10x=-12y$$

Combine -4x and -10x to get -14x.

$$-14x+16y=-12y$$

Add 12y to both sides.

$$-14x+16y+12y=0$$

Combine 16y and 12y to get 28y.

$$-14x+28y=0$$

Consider the second equation. Divide both sides by 20.

$$x-4y=\frac{-10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$x-4y=-\frac{1}{2}$$

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.

$$-14x+28y=0,x-4y=-\frac{1}{2}$$

To make -14x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by -14.

$$-14x+28y=0,-14x-14\left(-4\right)y=-14\left(-\frac{1}{2}\right)$$

Simplify.

$$-14x+28y=0,-14x+56y=7$$

Subtract -14x+56y=7 from -14x+28y=0 by subtracting like terms on each side of the equal sign.

$$-14x+14x+28y-56y=-7$$

Add -14x to 14x. Terms -14x and 14x cancel out, leaving an equation with only one variable that can be solved.

$$28y-56y=-7$$

Add 28y to -56y.

$$-28y=-7$$

Divide both sides by -28.

$$y=\frac{1}{4}$$

Substitute \frac{1}{4} for y in x-4y=-\frac{1}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-4\times \frac{1}{4}=-\frac{1}{2}$$

Multiply -4 times \frac{1}{4}.

$$x-1=-\frac{1}{2}$$

Add 1 to both sides of the equation.

$$x=\frac{1}{2}$$

The system is now solved.

$$x=\frac{1}{2},y=\frac{1}{4}$$