Solve {l}{75*y+30*x/5=3120}{100x+300y+520=3120}

$$\left.\begin{array} { l } { 75 \times y + 30 \cdot \frac { x } { 5 } = 3120 } \\ { 100 x + 300 y + 520 = 3120 } \end{array} \right\}$$

Solve for y, x

$$x=-130$$

$$y=52$$

Steps Using Substitution

Consider the first equation. Multiply both sides of the equation by 5.

$$375y+30x=15600$$

Consider the second equation. Subtract 520 from both sides.

$$100x+300y=3120-520$$

Subtract 520 from 3120 to get 2600.

$$100x+300y=2600$$

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.

$$375y+30x=15600,300y+100x=2600$$

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

$$375y+30x=15600$$

Subtract 30x from both sides of the equation.

$$375y=-30x+15600$$

Divide both sides by 375.

$$y=\frac{1}{375}\left(-30x+15600\right)$$

Multiply \frac{1}{375} times -30x+15600.

$$y=-\frac{2}{25}x+\frac{208}{5}$$

Substitute -\frac{2x}{25}+\frac{208}{5} for y in the other equation, 300y+100x=2600.

$$300\left(-\frac{2}{25}x+\frac{208}{5}\right)+100x=2600$$

Multiply 300 times -\frac{2x}{25}+\frac{208}{5}.

$$-24x+12480+100x=2600$$

Add -24x to 100x.

$$76x+12480=2600$$

Subtract 12480 from both sides of the equation.

$$76x=-9880$$

Divide both sides by 76.

$$x=-130$$

Substitute -130 for x in y=-\frac{2}{25}x+\frac{208}{5}. Because the resulting equation contains only one variable, you can solve for y directly.

$$y=-\frac{2}{25}\left(-130\right)+\frac{208}{5}$$

Multiply -\frac{2}{25} times -130.

$$y=\frac{52+208}{5}$$

Add \frac{208}{5} to \frac{52}{5} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$y=52$$

The system is now solved.

$$y=52,x=-130$$

Steps Using Matrices

Consider the first equation. Multiply both sides of the equation by 5.

$$375y+30x=15600$$

Consider the second equation. Subtract 520 from both sides.

$$100x+300y=3120-520$$

Subtract 520 from 3120 to get 2600.

$$100x+300y=2600$$

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

$$375y+30x=15600,300y+100x=2600$$

Write the equations in matrix form.

$$\left(\begin{matrix}375&30\\300&100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}375&30\\300&100\end{matrix}\right).

$$inverse(\left(\begin{matrix}375&30\\300&100\end{matrix}\right))\left(\begin{matrix}375&30\\300&100\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}375&30\\300&100\end{matrix}\right))\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}375&30\\300&100\end{matrix}\right))\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

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

$$\left(\begin{matrix}y\\x\end{matrix}\right)=inverse(\left(\begin{matrix}375&30\\300&100\end{matrix}\right))\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

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.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{100}{375\times 100-30\times 300}&-\frac{30}{375\times 100-30\times 300}\\-\frac{300}{375\times 100-30\times 300}&\frac{375}{375\times 100-30\times 300}\end{matrix}\right)\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{285}&-\frac{1}{950}\\-\frac{1}{95}&\frac{1}{76}\end{matrix}\right)\left(\begin{matrix}15600\\2600\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{285}\times 15600-\frac{1}{950}\times 2600\\-\frac{1}{95}\times 15600+\frac{1}{76}\times 2600\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}y\\x\end{matrix}\right)=\left(\begin{matrix}52\\-130\end{matrix}\right)$$

Extract the matrix elements y and x.

$$y=52,x=-130$$

Steps Using Elimination

Consider the first equation. Multiply both sides of the equation by 5.

$$375y+30x=15600$$

Consider the second equation. Subtract 520 from both sides.

$$100x+300y=3120-520$$

Subtract 520 from 3120 to get 2600.

$$100x+300y=2600$$

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.

$$375y+30x=15600,300y+100x=2600$$

To make 375y and 300y equal, multiply all terms on each side of the first equation by 300 and all terms on each side of the second by 375.

$$300\times 375y+300\times 30x=300\times 15600,375\times 300y+375\times 100x=375\times 2600$$

Simplify.

$$112500y+9000x=4680000,112500y+37500x=975000$$

Subtract 112500y+37500x=975000 from 112500y+9000x=4680000 by subtracting like terms on each side of the equal sign.

$$112500y-112500y+9000x-37500x=4680000-975000$$

Add 112500y to -112500y. Terms 112500y and -112500y cancel out, leaving an equation with only one variable that can be solved.

$$9000x-37500x=4680000-975000$$

Add 9000x to -37500x.

$$-28500x=4680000-975000$$

Add 4680000 to -975000.

$$-28500x=3705000$$

Divide both sides by -28500.

$$x=-130$$

Substitute -130 for x in 300y+100x=2600. Because the resulting equation contains only one variable, you can solve for y directly.

$$300y+100\left(-130\right)=2600$$

Multiply 100 times -130.

$$300y-13000=2600$$

Add 13000 to both sides of the equation.

$$300y=15600$$

Divide both sides by 300.

$$y=52$$

The system is now solved.

$$y=52,x=-130$$