Solve {l}{x+y=100}{60x+70y=630}

$$\left. \begin{array} { l } { x + y = 100 } \\ { 60 x + 70 y = 630 } \end{array} \right.$$

Solve for x, y

$$x=637$$

$$y=-537$$

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.

$$x+y=100,60x+70y=630$$

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

$$x+y=100$$

Subtract y from both sides of the equation.

$$x=-y+100$$

Substitute -y+100 for x in the other equation, 60x+70y=630.

$$60\left(-y+100\right)+70y=630$$

Multiply 60 times -y+100.

$$-60y+6000+70y=630$$

Add -60y to 70y.

$$10y+6000=630$$

Subtract 6000 from both sides of the equation.

$$10y=-5370$$

Divide both sides by 10.

$$y=-537$$

Substitute -537 for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\left(-537\right)+100$$

Multiply -1 times -537.

$$x=537+100$$

Add 100 to 537.

$$x=637$$

The system is now solved.

$$x=637,y=-537$$

Steps Using Matrices

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

$$x+y=100,60x+70y=630$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&1\\60&70\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\630\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\60&70\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&1\\60&70\end{matrix}\right))\left(\begin{matrix}1&1\\60&70\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\60&70\end{matrix}\right))\left(\begin{matrix}100\\630\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\60&70\end{matrix}\right))\left(\begin{matrix}100\\630\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\60&70\end{matrix}\right))\left(\begin{matrix}100\\630\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{70}{70-60}&-\frac{1}{70-60}\\-\frac{60}{70-60}&\frac{1}{70-60}\end{matrix}\right)\left(\begin{matrix}100\\630\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7&-\frac{1}{10}\\-6&\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}100\\630\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}7\times 100-\frac{1}{10}\times 630\\-6\times 100+\frac{1}{10}\times 630\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}637\\-537\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=637,y=-537$$

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.

$$x+y=100,60x+70y=630$$

To make x and 60x equal, multiply all terms on each side of the first equation by 60 and all terms on each side of the second by 1.

$$60x+60y=60\times 100,60x+70y=630$$

Simplify.

$$60x+60y=6000,60x+70y=630$$

Subtract 60x+70y=630 from 60x+60y=6000 by subtracting like terms on each side of the equal sign.

$$60x-60x+60y-70y=6000-630$$

Add 60x to -60x. Terms 60x and -60x cancel out, leaving an equation with only one variable that can be solved.

$$60y-70y=6000-630$$

Add 60y to -70y.

$$-10y=6000-630$$

Add 6000 to -630.

$$-10y=5370$$

Divide both sides by -10.

$$y=-537$$

Substitute -537 for y in 60x+70y=630. Because the resulting equation contains only one variable, you can solve for x directly.

$$60x+70\left(-537\right)=630$$

Multiply 70 times -537.

$$60x-37590=630$$

Add 37590 to both sides of the equation.

$$60x=38220$$

Divide both sides by 60.

$$x=637$$

The system is now solved.

$$x=637,y=-537$$