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x+y=70,2.5x+2y=189
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=70
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+70
Subtract y from both sides of the equation.
2.5\left(-y+70\right)+2y=189
Substitute -y+70 for x in the other equation, 2.5x+2y=189.
-2.5y+175+2y=189
Multiply 2.5 times -y+70.
-0.5y+175=189
Add -\frac{5y}{2} to 2y.
-0.5y=14
Subtract 175 from both sides of the equation.
y=-28
Multiply both sides by -2.
x=-\left(-28\right)+70
Substitute -28 for y in x=-y+70. Because the resulting equation contains only one variable, you can solve for x directly.
x=28+70
Multiply -1 times -28.
x=98
Add 70 to 28.
x=98,y=-28
The system is now solved.
x+y=70,2.5x+2y=189
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}70\\189\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right))\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right))\left(\begin{matrix}70\\189\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\2.5&2\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right))\left(\begin{matrix}70\\189\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\2.5&2\end{matrix}\right))\left(\begin{matrix}70\\189\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{2}{2-2.5}&-\frac{1}{2-2.5}\\-\frac{2.5}{2-2.5}&\frac{1}{2-2.5}\end{matrix}\right)\left(\begin{matrix}70\\189\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}-4&2\\5&-2\end{matrix}\right)\left(\begin{matrix}70\\189\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-4\times 70+2\times 189\\5\times 70-2\times 189\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}98\\-28\end{matrix}\right)
Do the arithmetic.
x=98,y=-28
Extract the matrix elements x and y.
x+y=70,2.5x+2y=189
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.
2.5x+2.5y=2.5\times 70,2.5x+2y=189
To make x and \frac{5x}{2} equal, multiply all terms on each side of the first equation by 2.5 and all terms on each side of the second by 1.
2.5x+2.5y=175,2.5x+2y=189
Simplify.
2.5x-2.5x+2.5y-2y=175-189
Subtract 2.5x+2y=189 from 2.5x+2.5y=175 by subtracting like terms on each side of the equal sign.
2.5y-2y=175-189
Add \frac{5x}{2} to -\frac{5x}{2}. Terms \frac{5x}{2} and -\frac{5x}{2} cancel out, leaving an equation with only one variable that can be solved.
0.5y=175-189
Add \frac{5y}{2} to -2y.
0.5y=-14
Add 175 to -189.
y=-28
Multiply both sides by 2.
2.5x+2\left(-28\right)=189
Substitute -28 for y in 2.5x+2y=189. Because the resulting equation contains only one variable, you can solve for x directly.
2.5x-56=189
Multiply 2 times -28.
2.5x=245
Add 56 to both sides of the equation.
x=98
Divide both sides of the equation by 2.5, which is the same as multiplying both sides by the reciprocal of the fraction.
x=98,y=-28
The system is now solved.