Solve {l}{5(2x+4)+3(4y-5)=41}{7(x-2)-6(8y-1)=130}

$$\left\{ \begin{array} { l } { 5 ( 2 x + 4 ) + 3 ( 4 y - 5 ) = 41 } \\ { 7 ( x - 2 ) - 6 ( 8 y - 1 ) = 130 } \end{array} \right.$$

Solve for x, y

$$x=6$$

$$y=-2$$

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.

$$5\left(2x+4\right)+3\left(4y-5\right)=41,7\left(x-2\right)-6\left(8y-1\right)=130$$

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

$$5\left(2x+4\right)+3\left(4y-5\right)=41$$

Multiply 5 times 4+2x.

$$10x+20+3\left(4y-5\right)=41$$

Multiply 3 times 4y-5.

$$10x+20+12y-15=41$$

Add 20 to -15.

$$10x+12y+5=41$$

Subtract 5 from both sides of the equation.

$$10x+12y=36$$

Subtract 12y from both sides of the equation.

$$10x=-12y+36$$

Divide both sides by 10.

$$x=\frac{1}{10}\left(-12y+36\right)$$

Multiply \frac{1}{10} times -12y+36.

$$x=-\frac{6}{5}y+\frac{18}{5}$$

Substitute \frac{-6y+18}{5} for x in the other equation, 7\left(x-2\right)-6\left(8y-1\right)=130.

$$7\left(-\frac{6}{5}y+\frac{18}{5}-2\right)-6\left(8y-1\right)=130$$

Add \frac{18}{5} to -2.

$$7\left(-\frac{6}{5}y+\frac{8}{5}\right)-6\left(8y-1\right)=130$$

Multiply 7 times \frac{-6y+8}{5}.

$$-\frac{42}{5}y+\frac{56}{5}-6\left(8y-1\right)=130$$

Multiply -6 times 8y-1.

$$-\frac{42}{5}y+\frac{56}{5}-48y+6=130$$

Add -\frac{42y}{5} to -48y.

$$-\frac{282}{5}y+\frac{56}{5}+6=130$$

Add \frac{56}{5} to 6.

$$-\frac{282}{5}y+\frac{86}{5}=130$$

Subtract \frac{86}{5} from both sides of the equation.

$$-\frac{282}{5}y=\frac{564}{5}$$

Divide both sides of the equation by -\frac{282}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-2$$

Substitute -2 for y in x=-\frac{6}{5}y+\frac{18}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{6}{5}\left(-2\right)+\frac{18}{5}$$

Multiply -\frac{6}{5} times -2.

$$x=\frac{12+18}{5}$$

Add \frac{18}{5} to \frac{12}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=6$$

The system is now solved.

$$x=6,y=-2$$

Steps Using Matrices

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

$$5\left(2x+4\right)+3\left(4y-5\right)=41,7\left(x-2\right)-6\left(8y-1\right)=130$$

Simplify the first equation to put it in standard form.

$$5\left(2x+4\right)+3\left(4y-5\right)=41$$

Multiply 5 times 2x+4.

$$10x+20+3\left(4y-5\right)=41$$

Multiply 3 times 4y-5.

$$10x+20+12y-15=41$$

Add 20 to -15.

$$10x+12y+5=41$$

Subtract 5 from both sides of the equation.

$$10x+12y=36$$

Simplify the second equation to put it in standard form.

$$7\left(x-2\right)-6\left(8y-1\right)=130$$

Multiply 7 times x-2.

$$7x-14-6\left(8y-1\right)=130$$

Multiply -6 times 8y-1.

$$7x-14-48y+6=130$$

Add -14 to 6.

$$7x-48y-8=130$$

Add 8 to both sides of the equation.

$$7x-48y=138$$

Write the equations in matrix form.

$$\left(\begin{matrix}10&12\\7&-48\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}36\\138\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&12\\7&-48\end{matrix}\right).

$$inverse(\left(\begin{matrix}10&12\\7&-48\end{matrix}\right))\left(\begin{matrix}10&12\\7&-48\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&12\\7&-48\end{matrix}\right))\left(\begin{matrix}36\\138\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&12\\7&-48\end{matrix}\right))\left(\begin{matrix}36\\138\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}10&12\\7&-48\end{matrix}\right))\left(\begin{matrix}36\\138\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{48}{10\left(-48\right)-12\times 7}&-\frac{12}{10\left(-48\right)-12\times 7}\\-\frac{7}{10\left(-48\right)-12\times 7}&\frac{10}{10\left(-48\right)-12\times 7}\end{matrix}\right)\left(\begin{matrix}36\\138\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{47}&\frac{1}{47}\\\frac{7}{564}&-\frac{5}{282}\end{matrix}\right)\left(\begin{matrix}36\\138\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{4}{47}\times 36+\frac{1}{47}\times 138\\\frac{7}{564}\times 36-\frac{5}{282}\times 138\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}6\\-2\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=6,y=-2$$