Solve {l}{x-3/5=y-7/2}{11x=13y}

$$\left. \begin{array} { l } { \frac { x - 3 } { 5 } = \frac { y - 7 } { 2 } } \\ { 11 x = 13 y } \end{array} \right.$$

Solve for x, y

$$x=13$$

$$y=11$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x-3\right)=5\left(y-7\right)$$

Use the distributive property to multiply 2 by x-3.

$$2x-6=5\left(y-7\right)$$

Use the distributive property to multiply 5 by y-7.

$$2x-6=5y-35$$

Subtract 5y from both sides.

$$2x-6-5y=-35$$

Add 6 to both sides.

$$2x-5y=-35+6$$

Add -35 and 6 to get -29.

$$2x-5y=-29$$

Consider the second equation. Subtract 13y from both sides.

$$11x-13y=0$$

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.

$$2x-5y=-29,11x-13y=0$$

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

$$2x-5y=-29$$

Add 5y to both sides of the equation.

$$2x=5y-29$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(5y-29\right)$$

Multiply \frac{1}{2} times 5y-29.

$$x=\frac{5}{2}y-\frac{29}{2}$$

Substitute \frac{5y-29}{2} for x in the other equation, 11x-13y=0.

$$11\left(\frac{5}{2}y-\frac{29}{2}\right)-13y=0$$

Multiply 11 times \frac{5y-29}{2}.

$$\frac{55}{2}y-\frac{319}{2}-13y=0$$

Add \frac{55y}{2} to -13y.

$$\frac{29}{2}y-\frac{319}{2}=0$$

Add \frac{319}{2} to both sides of the equation.

$$\frac{29}{2}y=\frac{319}{2}$$

Divide both sides of the equation by \frac{29}{2}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=11$$

Substitute 11 for y in x=\frac{5}{2}y-\frac{29}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{5}{2}\times 11-\frac{29}{2}$$

Multiply \frac{5}{2} times 11.

$$x=\frac{55-29}{2}$$

Add -\frac{29}{2} to \frac{55}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=13$$

The system is now solved.

$$x=13,y=11$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x-3\right)=5\left(y-7\right)$$

Use the distributive property to multiply 2 by x-3.

$$2x-6=5\left(y-7\right)$$

Use the distributive property to multiply 5 by y-7.

$$2x-6=5y-35$$

Subtract 5y from both sides.

$$2x-6-5y=-35$$

Add 6 to both sides.

$$2x-5y=-35+6$$

Add -35 and 6 to get -29.

$$2x-5y=-29$$

Consider the second equation. Subtract 13y from both sides.

$$11x-13y=0$$

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

$$2x-5y=-29,11x-13y=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-29\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-5\\11&-13\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right))\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right))\left(\begin{matrix}-29\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right))\left(\begin{matrix}-29\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-5\\11&-13\end{matrix}\right))\left(\begin{matrix}-29\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{2\left(-13\right)-\left(-5\times 11\right)}&-\frac{-5}{2\left(-13\right)-\left(-5\times 11\right)}\\-\frac{11}{2\left(-13\right)-\left(-5\times 11\right)}&\frac{2}{2\left(-13\right)-\left(-5\times 11\right)}\end{matrix}\right)\left(\begin{matrix}-29\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{29}&\frac{5}{29}\\-\frac{11}{29}&\frac{2}{29}\end{matrix}\right)\left(\begin{matrix}-29\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{13}{29}\left(-29\right)\\-\frac{11}{29}\left(-29\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}13\\11\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=13,y=11$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 10, the least common multiple of 5,2.

$$2\left(x-3\right)=5\left(y-7\right)$$

Use the distributive property to multiply 2 by x-3.

$$2x-6=5\left(y-7\right)$$

Use the distributive property to multiply 5 by y-7.

$$2x-6=5y-35$$

Subtract 5y from both sides.

$$2x-6-5y=-35$$

Add 6 to both sides.

$$2x-5y=-35+6$$

Add -35 and 6 to get -29.

$$2x-5y=-29$$

Consider the second equation. Subtract 13y from both sides.

$$11x-13y=0$$

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.

$$2x-5y=-29,11x-13y=0$$

To make 2x and 11x equal, multiply all terms on each side of the first equation by 11 and all terms on each side of the second by 2.

$$11\times 2x+11\left(-5\right)y=11\left(-29\right),2\times 11x+2\left(-13\right)y=0$$

Simplify.

$$22x-55y=-319,22x-26y=0$$

Subtract 22x-26y=0 from 22x-55y=-319 by subtracting like terms on each side of the equal sign.

$$22x-22x-55y+26y=-319$$

Add 22x to -22x. Terms 22x and -22x cancel out, leaving an equation with only one variable that can be solved.

$$-55y+26y=-319$$

Add -55y to 26y.

$$-29y=-319$$

Divide both sides by -29.

$$y=11$$

Substitute 11 for y in 11x-13y=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$11x-13\times 11=0$$

Multiply -13 times 11.

$$11x-143=0$$

Add 143 to both sides of the equation.

$$11x=143$$

Divide both sides by 11.

$$x=13$$

The system is now solved.

$$x=13,y=11$$