Solve {l}{8x-5y=49}{7x+15y=101}

$$\left\{ \begin{array} { l } { 8 x - 5 y = 49 } \\ { 7 x + 15 y = 101 } \end{array} \right.$$

Solve for x, y

$$x=8$$

$$y=3$$

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.

$$8x-5y=49,7x+15y=101$$

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

$$8x-5y=49$$

Add 5y to both sides of the equation.

$$8x=5y+49$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(5y+49\right)$$

Multiply \frac{1}{8} times 5y+49.

$$x=\frac{5}{8}y+\frac{49}{8}$$

Substitute \frac{5y+49}{8} for x in the other equation, 7x+15y=101.

$$7\left(\frac{5}{8}y+\frac{49}{8}\right)+15y=101$$

Multiply 7 times \frac{5y+49}{8}.

$$\frac{35}{8}y+\frac{343}{8}+15y=101$$

Add \frac{35y}{8} to 15y.

$$\frac{155}{8}y+\frac{343}{8}=101$$

Subtract \frac{343}{8} from both sides of the equation.

$$\frac{155}{8}y=\frac{465}{8}$$

Divide both sides of the equation by \frac{155}{8}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=3$$

Substitute 3 for y in x=\frac{5}{8}y+\frac{49}{8}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{8}\times 3+\frac{49}{8}$$

Multiply \frac{5}{8} times 3.

$$x=\frac{15+49}{8}$$

Add \frac{49}{8} to \frac{15}{8} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=8$$

The system is now solved.

$$x=8,y=3$$

Steps Using Matrices

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

$$8x-5y=49,7x+15y=101$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&-5\\7&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}49\\101\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&-5\\7&15\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&-5\\7&15\end{matrix}\right))\left(\begin{matrix}8&-5\\7&15\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-5\\7&15\end{matrix}\right))\left(\begin{matrix}49\\101\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-5\\7&15\end{matrix}\right))\left(\begin{matrix}49\\101\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&-5\\7&15\end{matrix}\right))\left(\begin{matrix}49\\101\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{15}{8\times 15-\left(-5\times 7\right)}&-\frac{-5}{8\times 15-\left(-5\times 7\right)}\\-\frac{7}{8\times 15-\left(-5\times 7\right)}&\frac{8}{8\times 15-\left(-5\times 7\right)}\end{matrix}\right)\left(\begin{matrix}49\\101\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{31}&\frac{1}{31}\\-\frac{7}{155}&\frac{8}{155}\end{matrix}\right)\left(\begin{matrix}49\\101\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{3}{31}\times 49+\frac{1}{31}\times 101\\-\frac{7}{155}\times 49+\frac{8}{155}\times 101\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}8\\3\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=8,y=3$$

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.

$$8x-5y=49,7x+15y=101$$

To make 8x 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 8.

$$7\times 8x+7\left(-5\right)y=7\times 49,8\times 7x+8\times 15y=8\times 101$$

Simplify.

$$56x-35y=343,56x+120y=808$$

Subtract 56x+120y=808 from 56x-35y=343 by subtracting like terms on each side of the equal sign.

$$56x-56x-35y-120y=343-808$$

Add 56x to -56x. Terms 56x and -56x cancel out, leaving an equation with only one variable that can be solved.

$$-35y-120y=343-808$$

Add -35y to -120y.

$$-155y=343-808$$

Add 343 to -808.

$$-155y=-465$$

Divide both sides by -155.

$$y=3$$

Substitute 3 for y in 7x+15y=101. Because the resulting equation contains only one variable, you can solve for x directly.

$$7x+15\times 3=101$$

Multiply 15 times 3.

$$7x+45=101$$

Subtract 45 from both sides of the equation.

$$7x=56$$

Divide both sides by 7.

$$x=8$$

The system is now solved.

$$x=8,y=3$$