Solve {l}{5x+7y=2060}{1020=2060-(2x+4y)}

$$\left\{ \begin{array} { l } { 5 x + 7 y = 2060 } \\ { 1020 = 2060 - ( 2 x + 4 y ) } \end{array} \right.$$

Solve for x, y

$$x=160$$

$$y=180$$

Steps Using Substitution

Consider the second equation. To find the opposite of 2x+4y, find the opposite of each term.

$$1020=2060-2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2060-2x-4y=1020$$

Subtract 2060 from both sides.

$$-2x-4y=1020-2060$$

Subtract 2060 from 1020 to get -1040.

$$-2x-4y=-1040$$

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.

$$5x+7y=2060,-2x-4y=-1040$$

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

$$5x+7y=2060$$

Subtract 7y from both sides of the equation.

$$5x=-7y+2060$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(-7y+2060\right)$$

Multiply \frac{1}{5} times -7y+2060.

$$x=-\frac{7}{5}y+412$$

Substitute -\frac{7y}{5}+412 for x in the other equation, -2x-4y=-1040.

$$-2\left(-\frac{7}{5}y+412\right)-4y=-1040$$

Multiply -2 times -\frac{7y}{5}+412.

$$\frac{14}{5}y-824-4y=-1040$$

Add \frac{14y}{5} to -4y.

$$-\frac{6}{5}y-824=-1040$$

Add 824 to both sides of the equation.

$$-\frac{6}{5}y=-216$$

Divide both sides of the equation by -\frac{6}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=180$$

Substitute 180 for y in x=-\frac{7}{5}y+412. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{7}{5}\times 180+412$$

Multiply -\frac{7}{5} times 180.

$$x=-252+412$$

Add 412 to -252.

$$x=160$$

The system is now solved.

$$x=160,y=180$$

Steps Using Matrices

Consider the second equation. To find the opposite of 2x+4y, find the opposite of each term.

$$1020=2060-2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2060-2x-4y=1020$$

Subtract 2060 from both sides.

$$-2x-4y=1020-2060$$

Subtract 2060 from 1020 to get -1040.

$$-2x-4y=-1040$$

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

$$5x+7y=2060,-2x-4y=-1040$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&7\\-2&-4\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right))\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right))\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right))\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&7\\-2&-4\end{matrix}\right))\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{4}{5\left(-4\right)-7\left(-2\right)}&-\frac{7}{5\left(-4\right)-7\left(-2\right)}\\-\frac{-2}{5\left(-4\right)-7\left(-2\right)}&\frac{5}{5\left(-4\right)-7\left(-2\right)}\end{matrix}\right)\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}&\frac{7}{6}\\-\frac{1}{3}&-\frac{5}{6}\end{matrix}\right)\left(\begin{matrix}2060\\-1040\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{3}\times 2060+\frac{7}{6}\left(-1040\right)\\-\frac{1}{3}\times 2060-\frac{5}{6}\left(-1040\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}160\\180\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=160,y=180$$

Steps Using Elimination

Consider the second equation. To find the opposite of 2x+4y, find the opposite of each term.

$$1020=2060-2x-4y$$

Swap sides so that all variable terms are on the left hand side.

$$2060-2x-4y=1020$$

Subtract 2060 from both sides.

$$-2x-4y=1020-2060$$

Subtract 2060 from 1020 to get -1040.

$$-2x-4y=-1040$$

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.

$$5x+7y=2060,-2x-4y=-1040$$

To make 5x and -2x equal, multiply all terms on each side of the first equation by -2 and all terms on each side of the second by 5.

$$-2\times 5x-2\times 7y=-2\times 2060,5\left(-2\right)x+5\left(-4\right)y=5\left(-1040\right)$$

Simplify.

$$-10x-14y=-4120,-10x-20y=-5200$$

Subtract -10x-20y=-5200 from -10x-14y=-4120 by subtracting like terms on each side of the equal sign.

$$-10x+10x-14y+20y=-4120+5200$$

Add -10x to 10x. Terms -10x and 10x cancel out, leaving an equation with only one variable that can be solved.

$$-14y+20y=-4120+5200$$

Add -14y to 20y.

$$6y=-4120+5200$$

Add -4120 to 5200.

$$6y=1080$$

Divide both sides by 6.

$$y=180$$

Substitute 180 for y in -2x-4y=-1040. Because the resulting equation contains only one variable, you can solve for x directly.

$$-2x-4\times 180=-1040$$

Multiply -4 times 180.

$$-2x-720=-1040$$

Add 720 to both sides of the equation.

$$-2x=-320$$

Divide both sides by -2.

$$x=160$$

The system is now solved.

$$x=160,y=180$$