Solve {l}{2x/15+7y/12=3}{7x/25-5y/16=frac{3}{20}}

$$\left. \begin{array} { l } { \frac { 2 x } { 15 } + \frac { 7 y } { 12 } = 3 } \\ { \frac { 7 x } { 25 } - \frac { 5 y } { 16 } = \frac { 3 } { 20 } } \end{array} \right.$$

Solve for x, y

$$x=5$$

$$y=4$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 15,12.

$$4\times 2x+5\times 7y=180$$

Multiply 4 and 2 to get 8.

$$8x+5\times 7y=180$$

Multiply 5 and 7 to get 35.

$$8x+35y=180$$

Consider the second equation. Multiply both sides of the equation by 400, the least common multiple of 25,16,20.

$$16\times 7x-25\times 5y=60$$

Multiply 16 and 7 to get 112.

$$112x-25\times 5y=60$$

Multiply -25 and 5 to get -125.

$$112x-125y=60$$

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.

$$8x+35y=180,112x-125y=60$$

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

$$8x+35y=180$$

Subtract 35y from both sides of the equation.

$$8x=-35y+180$$

Divide both sides by 8.

$$x=\frac{1}{8}\left(-35y+180\right)$$

Multiply \frac{1}{8} times -35y+180.

$$x=-\frac{35}{8}y+\frac{45}{2}$$

Substitute -\frac{35y}{8}+\frac{45}{2} for x in the other equation, 112x-125y=60.

$$112\left(-\frac{35}{8}y+\frac{45}{2}\right)-125y=60$$

Multiply 112 times -\frac{35y}{8}+\frac{45}{2}.

$$-490y+2520-125y=60$$

Add -490y to -125y.

$$-615y+2520=60$$

Subtract 2520 from both sides of the equation.

$$-615y=-2460$$

Divide both sides by -615.

$$y=4$$

Substitute 4 for y in x=-\frac{35}{8}y+\frac{45}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{35}{8}\times 4+\frac{45}{2}$$

Multiply -\frac{35}{8} times 4.

$$x=\frac{-35+45}{2}$$

Add \frac{45}{2} to -\frac{35}{2} 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.

$$x=5,y=4$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 15,12.

$$4\times 2x+5\times 7y=180$$

Multiply 4 and 2 to get 8.

$$8x+5\times 7y=180$$

Multiply 5 and 7 to get 35.

$$8x+35y=180$$

Consider the second equation. Multiply both sides of the equation by 400, the least common multiple of 25,16,20.

$$16\times 7x-25\times 5y=60$$

Multiply 16 and 7 to get 112.

$$112x-25\times 5y=60$$

Multiply -25 and 5 to get -125.

$$112x-125y=60$$

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

$$8x+35y=180,112x-125y=60$$

Write the equations in matrix form.

$$\left(\begin{matrix}8&35\\112&-125\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}180\\60\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}8&35\\112&-125\end{matrix}\right).

$$inverse(\left(\begin{matrix}8&35\\112&-125\end{matrix}\right))\left(\begin{matrix}8&35\\112&-125\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&35\\112&-125\end{matrix}\right))\left(\begin{matrix}180\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&35\\112&-125\end{matrix}\right))\left(\begin{matrix}180\\60\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}8&35\\112&-125\end{matrix}\right))\left(\begin{matrix}180\\60\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{125}{8\left(-125\right)-35\times 112}&-\frac{35}{8\left(-125\right)-35\times 112}\\-\frac{112}{8\left(-125\right)-35\times 112}&\frac{8}{8\left(-125\right)-35\times 112}\end{matrix}\right)\left(\begin{matrix}180\\60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{984}&\frac{7}{984}\\\frac{14}{615}&-\frac{1}{615}\end{matrix}\right)\left(\begin{matrix}180\\60\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{25}{984}\times 180+\frac{7}{984}\times 60\\\frac{14}{615}\times 180-\frac{1}{615}\times 60\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}5\\4\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=5,y=4$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 60, the least common multiple of 15,12.

$$4\times 2x+5\times 7y=180$$

Multiply 4 and 2 to get 8.

$$8x+5\times 7y=180$$

Multiply 5 and 7 to get 35.

$$8x+35y=180$$

Consider the second equation. Multiply both sides of the equation by 400, the least common multiple of 25,16,20.

$$16\times 7x-25\times 5y=60$$

Multiply 16 and 7 to get 112.

$$112x-25\times 5y=60$$

Multiply -25 and 5 to get -125.

$$112x-125y=60$$

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.

$$8x+35y=180,112x-125y=60$$

To make 8x and 112x equal, multiply all terms on each side of the first equation by 112 and all terms on each side of the second by 8.

$$112\times 8x+112\times 35y=112\times 180,8\times 112x+8\left(-125\right)y=8\times 60$$

Simplify.

$$896x+3920y=20160,896x-1000y=480$$

Subtract 896x-1000y=480 from 896x+3920y=20160 by subtracting like terms on each side of the equal sign.

$$896x-896x+3920y+1000y=20160-480$$

Add 896x to -896x. Terms 896x and -896x cancel out, leaving an equation with only one variable that can be solved.

$$3920y+1000y=20160-480$$

Add 3920y to 1000y.

$$4920y=20160-480$$

Add 20160 to -480.

$$4920y=19680$$

Divide both sides by 4920.

$$y=4$$

Substitute 4 for y in 112x-125y=60. Because the resulting equation contains only one variable, you can solve for x directly.

$$112x-125\times 4=60$$

Multiply -125 times 4.

$$112x-500=60$$

Add 500 to both sides of the equation.

$$112x=560$$

Divide both sides by 112.

$$x=5$$

The system is now solved.

$$x=5,y=4$$