Solve {r}{5(x+5)-2(y-3)=62}{4(x-7)-(y+4)=-9}

$$\left. \begin{array} { r } { 5 ( x + 5 ) - 2 ( y - 3 ) = 62 } \\ { 4 ( x - 7 ) - ( y + 4 ) = - 9 } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=-3$$

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.

$$5\left(x+5\right)-2\left(y-3\right)=62,4\left(x-7\right)-\left(y+4\right)=-9$$

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

$$5\left(x+5\right)-2\left(y-3\right)=62$$

Multiply 5 times x+5.

$$5x+25-2\left(y-3\right)=62$$

Multiply -2 times y-3.

$$5x+25-2y+6=62$$

Add 25 to 6.

$$5x-2y+31=62$$

Subtract 31 from both sides of the equation.

$$5x-2y=31$$

Add 2y to both sides of the equation.

$$5x=2y+31$$

Divide both sides by 5.

$$x=\frac{1}{5}\left(2y+31\right)$$

Multiply \frac{1}{5} times 2y+31.

$$x=\frac{2}{5}y+\frac{31}{5}$$

Substitute \frac{2y+31}{5} for x in the other equation, 4\left(x-7\right)-\left(y+4\right)=-9.

$$4\left(\frac{2}{5}y+\frac{31}{5}-7\right)-\left(y+4\right)=-9$$

Add \frac{31}{5} to -7.

$$4\left(\frac{2}{5}y-\frac{4}{5}\right)-\left(y+4\right)=-9$$

Multiply 4 times \frac{-4+2y}{5}.

$$\frac{8}{5}y-\frac{16}{5}-\left(y+4\right)=-9$$

Multiply -1 times y+4.

$$\frac{8}{5}y-\frac{16}{5}-y-4=-9$$

Add \frac{8y}{5} to -y.

$$\frac{3}{5}y-\frac{16}{5}-4=-9$$

Add -\frac{16}{5} to -4.

$$\frac{3}{5}y-\frac{36}{5}=-9$$

Add \frac{36}{5} to both sides of the equation.

$$\frac{3}{5}y=-\frac{9}{5}$$

Divide both sides of the equation by \frac{3}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=-3$$

Substitute -3 for y in x=\frac{2}{5}y+\frac{31}{5}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{2}{5}\left(-3\right)+\frac{31}{5}$$

Multiply \frac{2}{5} times -3.

$$x=\frac{-6+31}{5}$$

Add \frac{31}{5} to -\frac{6}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=5$$

The system is now solved.

$$x=5,y=-3$$

Steps Using Matrices

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

$$5\left(x+5\right)-2\left(y-3\right)=62,4\left(x-7\right)-\left(y+4\right)=-9$$

Simplify the first equation to put it in standard form.

$$5\left(x+5\right)-2\left(y-3\right)=62$$

Multiply 5 times x+5.

$$5x+25-2\left(y-3\right)=62$$

Multiply -2 times y-3.

$$5x+25-2y+6=62$$

Add 25 to 6.

$$5x-2y+31=62$$

Subtract 31 from both sides of the equation.

$$5x-2y=31$$

Simplify the second equation to put it in standard form.

$$4\left(x-7\right)-\left(y+4\right)=-9$$

Multiply 4 times x-7.

$$4x-28-\left(y+4\right)=-9$$

Multiply -1 times y+4.

$$4x-28-y-4=-9$$

Add -28 to -4.

$$4x-y-32=-9$$

Add 32 to both sides of the equation.

$$4x-y=23$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}31\\23\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-2\\4&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right))\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right))\left(\begin{matrix}31\\23\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right))\left(\begin{matrix}31\\23\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}5&-2\\4&-1\end{matrix}\right))\left(\begin{matrix}31\\23\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5\left(-1\right)-\left(-2\times 4\right)}&-\frac{-2}{5\left(-1\right)-\left(-2\times 4\right)}\\-\frac{4}{5\left(-1\right)-\left(-2\times 4\right)}&\frac{5}{5\left(-1\right)-\left(-2\times 4\right)}\end{matrix}\right)\left(\begin{matrix}31\\23\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}&\frac{2}{3}\\-\frac{4}{3}&\frac{5}{3}\end{matrix}\right)\left(\begin{matrix}31\\23\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{3}\times 31+\frac{2}{3}\times 23\\-\frac{4}{3}\times 31+\frac{5}{3}\times 23\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\-3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=-3$$