Solve {l}{2x+30y=36}{30x+20y=34}

$$\left\{ \begin{array} { l } { 2 x + 30 y = 36 } \\ { 30 x + 20 y = 34 } \end{array} \right.$$

Solve for x, y

$$x=\frac{15}{43}\approx 0.348837209$$

$$y = \frac{253}{215} = 1\frac{38}{215} \approx 1.176744186$$

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.

$$2x+30y=36,30x+20y=34$$

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

$$2x+30y=36$$

Subtract 30y from both sides of the equation.

$$2x=-30y+36$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(-30y+36\right)$$

Multiply \frac{1}{2} times -30y+36.

$$x=-15y+18$$

Substitute -15y+18 for x in the other equation, 30x+20y=34.

$$30\left(-15y+18\right)+20y=34$$

Multiply 30 times -15y+18.

$$-450y+540+20y=34$$

Add -450y to 20y.

$$-430y+540=34$$

Subtract 540 from both sides of the equation.

$$-430y=-506$$

Divide both sides by -430.

$$y=\frac{253}{215}$$

Substitute \frac{253}{215} for y in x=-15y+18. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-15\times \frac{253}{215}+18$$

Multiply -15 times \frac{253}{215}.

$$x=-\frac{759}{43}+18$$

Add 18 to -\frac{759}{43}.

$$x=\frac{15}{43}$$

The system is now solved.

$$x=\frac{15}{43},y=\frac{253}{215}$$

Steps Using Matrices

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

$$2x+30y=36,30x+20y=34$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&30\\30&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\34\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&30\\30&20\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&30\\30&20\end{matrix}\right))\left(\begin{matrix}2&30\\30&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&30\\30&20\end{matrix}\right))\left(\begin{matrix}36\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&30\\30&20\end{matrix}\right))\left(\begin{matrix}36\\34\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&30\\30&20\end{matrix}\right))\left(\begin{matrix}36\\34\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{2\times 20-30\times 30}&-\frac{30}{2\times 20-30\times 30}\\-\frac{30}{2\times 20-30\times 30}&\frac{2}{2\times 20-30\times 30}\end{matrix}\right)\left(\begin{matrix}36\\34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{43}&\frac{3}{86}\\\frac{3}{86}&-\frac{1}{430}\end{matrix}\right)\left(\begin{matrix}36\\34\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{43}\times 36+\frac{3}{86}\times 34\\\frac{3}{86}\times 36-\frac{1}{430}\times 34\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{43}\\\frac{253}{215}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{15}{43},y=\frac{253}{215}$$

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.

$$2x+30y=36,30x+20y=34$$

To make 2x and 30x equal, multiply all terms on each side of the first equation by 30 and all terms on each side of the second by 2.

$$30\times 2x+30\times 30y=30\times 36,2\times 30x+2\times 20y=2\times 34$$

Simplify.

$$60x+900y=1080,60x+40y=68$$

Subtract 60x+40y=68 from 60x+900y=1080 by subtracting like terms on each side of the equal sign.

$$60x-60x+900y-40y=1080-68$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$900y-40y=1080-68$$

Add 900y to -40y.

$$860y=1080-68$$

Add 1080 to -68.

$$860y=1012$$

Divide both sides by 860.

$$y=\frac{253}{215}$$

Substitute \frac{253}{215} for y in 30x+20y=34. Because the resulting equation contains only one variable, you can solve for x directly.

$$30x+20\times \frac{253}{215}=34$$

Multiply 20 times \frac{253}{215}.

$$30x+\frac{1012}{43}=34$$

Subtract \frac{1012}{43} from both sides of the equation.

$$30x=\frac{450}{43}$$

Divide both sides by 30.

$$x=\frac{15}{43}$$

The system is now solved.

$$x=\frac{15}{43},y=\frac{253}{215}$$