Solve {l}{5x+y=2y*8=}{7x+y=25.85}

$$\left. \begin{array} { l } { 5 x + y = 2 y \cdot 8 = } \\ { 7 x + y = 25.85 } \end{array} \right.$$

Solve for x, y

$$x=3.525$$

$$y=1.175$$

Steps Using Substitution

Consider the first equation. Multiply 2 and 8 to get 16.

$$5x+y=16y$$

Subtract 16y from both sides.

$$5x+y-16y=0$$

Combine y and -16y to get -15y.

$$5x-15y=0$$

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.

$$5x-15y=0,7x+y=25.85$$

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

$$5x-15y=0$$

Add 15y to both sides of the equation.

$$5x=15y$$

Divide both sides by 5.

$$x=\frac{1}{5}\times 15y$$

Multiply \frac{1}{5} times 15y.

$$x=3y$$

Substitute 3y for x in the other equation, 7x+y=25.85.

$$7\times 3y+y=25.85$$

Multiply 7 times 3y.

$$21y+y=25.85$$

Add 21y to y.

$$22y=25.85$$

Divide both sides by 22.

$$y=1.175$$

Substitute 1.175 for y in x=3y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=3\times 1.175$$

Multiply 3 times 1.175.

$$x=3.525$$

The system is now solved.

$$x=3.525,y=1.175$$

Steps Using Matrices

Consider the first equation. Multiply 2 and 8 to get 16.

$$5x+y=16y$$

Subtract 16y from both sides.

$$5x+y-16y=0$$

Combine y and -16y to get -15y.

$$5x-15y=0$$

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

$$5x-15y=0,7x+y=25.85$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-15\\7&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-15\\7&1\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-15\\7&1\end{matrix}\right))\left(\begin{matrix}5&-15\\7&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-15\\7&1\end{matrix}\right))\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-15\\7&1\end{matrix}\right))\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-15\\7&1\end{matrix}\right))\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{5-\left(-15\times 7\right)}&-\frac{-15}{5-\left(-15\times 7\right)}\\-\frac{7}{5-\left(-15\times 7\right)}&\frac{5}{5-\left(-15\times 7\right)}\end{matrix}\right)\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{110}&\frac{3}{22}\\-\frac{7}{110}&\frac{1}{22}\end{matrix}\right)\left(\begin{matrix}0\\25.85\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{22}\times 25.85\\\frac{1}{22}\times 25.85\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{141}{40}\\\frac{47}{40}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{141}{40},y=\frac{47}{40}$$

Steps Using Elimination

Consider the first equation. Multiply 2 and 8 to get 16.

$$5x+y=16y$$

Subtract 16y from both sides.

$$5x+y-16y=0$$

Combine y and -16y to get -15y.

$$5x-15y=0$$

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.

$$5x-15y=0,7x+y=25.85$$

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.

$$7\times 5x+7\left(-15\right)y=0,5\times 7x+5y=5\times 25.85$$

Simplify.

$$35x-105y=0,35x+5y=129.25$$

Subtract 35x+5y=129.25 from 35x-105y=0 by subtracting like terms on each side of the equal sign.

$$35x-35x-105y-5y=-129.25$$

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

$$-105y-5y=-129.25$$

Add -105y to -5y.

$$-110y=-129.25$$

Divide both sides by -110.

$$y=\frac{47}{40}$$

Substitute \frac{47}{40} for y in 7x+y=25.85. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x+\frac{47}{40}=25.85$$

Subtract \frac{47}{40} from both sides of the equation.

$$7x=\frac{987}{40}$$

Divide both sides by 7.

$$x=\frac{141}{40}$$

The system is now solved.

$$x=\frac{141}{40},y=\frac{47}{40}$$