Solve {l}{30x+15y=675}{42x+20y=940}

$$\left\{ \begin{array} { l } { 30 x + 15 y = 675 } \\ { 42 x + 20 y = 940 } \end{array} \right.$$

Solve for x, y

$$x=20$$

$$y=5$$

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.

$$30x+15y=675,42x+20y=940$$

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

$$30x+15y=675$$

Subtract 15y from both sides of the equation.

$$30x=-15y+675$$

Divide both sides by 30.

$$x=\frac{1}{30}\left(-15y+675\right)$$

Multiply \frac{1}{30} times -15y+675.

$$x=-\frac{1}{2}y+\frac{45}{2}$$

Substitute \frac{-y+45}{2} for x in the other equation, 42x+20y=940.

$$42\left(-\frac{1}{2}y+\frac{45}{2}\right)+20y=940$$

Multiply 42 times \frac{-y+45}{2}.

$$-21y+945+20y=940$$

Add -21y to 20y.

$$-y+945=940$$

Subtract 945 from both sides of the equation.

$$-y=-5$$

Divide both sides by -1.

$$y=5$$

Substitute 5 for y in x=-\frac{1}{2}y+\frac{45}{2}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{2}\times 5+\frac{45}{2}$$

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

$$x=\frac{-5+45}{2}$$

Add \frac{45}{2} to -\frac{5}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=20$$

The system is now solved.

$$x=20,y=5$$

Steps Using Matrices

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

$$30x+15y=675,42x+20y=940$$

Write the equations in matrix form.

$$\left(\begin{matrix}30&15\\42&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}675\\940\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}30&15\\42&20\end{matrix}\right).

$$inverse(\left(\begin{matrix}30&15\\42&20\end{matrix}\right))\left(\begin{matrix}30&15\\42&20\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&15\\42&20\end{matrix}\right))\left(\begin{matrix}675\\940\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&15\\42&20\end{matrix}\right))\left(\begin{matrix}675\\940\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}30&15\\42&20\end{matrix}\right))\left(\begin{matrix}675\\940\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{20}{30\times 20-15\times 42}&-\frac{15}{30\times 20-15\times 42}\\-\frac{42}{30\times 20-15\times 42}&\frac{30}{30\times 20-15\times 42}\end{matrix}\right)\left(\begin{matrix}675\\940\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}&\frac{1}{2}\\\frac{7}{5}&-1\end{matrix}\right)\left(\begin{matrix}675\\940\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{2}{3}\times 675+\frac{1}{2}\times 940\\\frac{7}{5}\times 675-940\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}20\\5\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=20,y=5$$

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.

$$30x+15y=675,42x+20y=940$$

To make 30x and 42x equal, multiply all terms on each side of the first equation by 42 and all terms on each side of the second by 30.

$$42\times 30x+42\times 15y=42\times 675,30\times 42x+30\times 20y=30\times 940$$

Simplify.

$$1260x+630y=28350,1260x+600y=28200$$

Subtract 1260x+600y=28200 from 1260x+630y=28350 by subtracting like terms on each side of the equal sign.

$$1260x-1260x+630y-600y=28350-28200$$

Add 1260x to -1260x. Terms 1260x and -1260x cancel out, leaving an equation with only one variable that can be solved.

$$630y-600y=28350-28200$$

Add 630y to -600y.

$$30y=28350-28200$$

Add 28350 to -28200.

$$30y=150$$

Divide both sides by 30.

$$y=5$$

Substitute 5 for y in 42x+20y=940. Because the resulting equation contains only one variable, you can solve for x directly.

$$42x+20\times 5=940$$

Multiply 20 times 5.

$$42x+100=940$$

Subtract 100 from both sides of the equation.

$$42x=840$$

Divide both sides by 42.

$$x=20$$

The system is now solved.

$$x=20,y=5$$