Solve {l}{35=x+y}{315=12x+7y}

$$\left. \begin{array} { l } { 35 = x + y } \\ { 315 = 12 x + 7 y } \end{array} \right.$$

Solve for x, y

$$x=14$$

$$y=21$$

Steps Using Substitution

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$x+y=35$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$12x+7y=315$$

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.

$$x+y=35,12x+7y=315$$

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

$$x+y=35$$

Subtract y from both sides of the equation.

$$x=-y+35$$

Substitute -y+35 for x in the other equation, 12x+7y=315.

$$12\left(-y+35\right)+7y=315$$

Multiply 12 times -y+35.

$$-12y+420+7y=315$$

Add -12y to 7y.

$$-5y+420=315$$

Subtract 420 from both sides of the equation.

$$-5y=-105$$

Divide both sides by -5.

$$y=21$$

Substitute 21 for y in x=-y+35. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-21+35$$

Add 35 to -21.

$$x=14$$

The system is now solved.

$$x=14,y=21$$

Steps Using Matrices

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$x+y=35$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$12x+7y=315$$

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

$$x+y=35,12x+7y=315$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\12&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35\\315\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\12&7\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\12&7\end{matrix}\right))\left(\begin{matrix}1&1\\12&7\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12&7\end{matrix}\right))\left(\begin{matrix}35\\315\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12&7\end{matrix}\right))\left(\begin{matrix}35\\315\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\12&7\end{matrix}\right))\left(\begin{matrix}35\\315\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7}{7-12}&-\frac{1}{7-12}\\-\frac{12}{7-12}&\frac{1}{7-12}\end{matrix}\right)\left(\begin{matrix}35\\315\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{5}&\frac{1}{5}\\\frac{12}{5}&-\frac{1}{5}\end{matrix}\right)\left(\begin{matrix}35\\315\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{7}{5}\times 35+\frac{1}{5}\times 315\\\frac{12}{5}\times 35-\frac{1}{5}\times 315\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}14\\21\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=14,y=21$$

Steps Using Elimination

Consider the first equation. Swap sides so that all variable terms are on the left hand side.

$$x+y=35$$

Consider the second equation. Swap sides so that all variable terms are on the left hand side.

$$12x+7y=315$$

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.

$$x+y=35,12x+7y=315$$

To make x and 12x equal, multiply all terms on each side of the first equation by 12 and all terms on each side of the second by 1.

$$12x+12y=12\times 35,12x+7y=315$$

Simplify.

$$12x+12y=420,12x+7y=315$$

Subtract 12x+7y=315 from 12x+12y=420 by subtracting like terms on each side of the equal sign.

$$12x-12x+12y-7y=420-315$$

Add 12x to -12x. Terms 12x and -12x cancel out, leaving an equation with only one variable that can be solved.

$$12y-7y=420-315$$

Add 12y to -7y.

$$5y=420-315$$

Add 420 to -315.

$$5y=105$$

Divide both sides by 5.

$$y=21$$

Substitute 21 for y in 12x+7y=315. Because the resulting equation contains only one variable, you can solve for x directly.

$$12x+7\times 21=315$$

Multiply 7 times 21.

$$12x+147=315$$

Subtract 147 from both sides of the equation.

$$12x=168$$

Divide both sides by 12.

$$x=14$$

The system is now solved.

$$x=14,y=21$$