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0.16x+0.2y=60,x+y=15
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.
0.16x+0.2y=60
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
0.16x=-0.2y+60
Subtract \frac{y}{5} from both sides of the equation.
x=6.25\left(-0.2y+60\right)
Divide both sides of the equation by 0.16, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-1.25y+375
Multiply 6.25 times -\frac{y}{5}+60.
-1.25y+375+y=15
Substitute -\frac{5y}{4}+375 for x in the other equation, x+y=15.
-0.25y+375=15
Add -\frac{5y}{4} to y.
-0.25y=-360
Subtract 375 from both sides of the equation.
y=1440
Multiply both sides by -4.
x=-1.25\times 1440+375
Substitute 1440 for y in x=-1.25y+375. Because the resulting equation contains only one variable, you can solve for x directly.
x=-1800+375
Multiply -1.25 times 1440.
x=-1425
Add 375 to -1800.
x=-1425,y=1440
The system is now solved.
0.16x+0.2y=60,x+y=15
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}0.16&0.2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}60\\15\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}0.16&0.2\\1&1\end{matrix}\right))\left(\begin{matrix}0.16&0.2\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}0.16&0.2\\1&1\end{matrix}\right))\left(\begin{matrix}60\\15\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}0.16&0.2\\1&1\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}0.16&0.2\\1&1\end{matrix}\right))\left(\begin{matrix}60\\15\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}0.16&0.2\\1&1\end{matrix}\right))\left(\begin{matrix}60\\15\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{1}{0.16-0.2}&-\frac{0.2}{0.16-0.2}\\-\frac{1}{0.16-0.2}&\frac{0.16}{0.16-0.2}\end{matrix}\right)\left(\begin{matrix}60\\15\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}-25&5\\25&-4\end{matrix}\right)\left(\begin{matrix}60\\15\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-25\times 60+5\times 15\\25\times 60-4\times 15\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1425\\1440\end{matrix}\right)
Do the arithmetic.
x=-1425,y=1440
Extract the matrix elements x and y.
0.16x+0.2y=60,x+y=15
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.
0.16x+0.2y=60,0.16x+0.16y=0.16\times 15
To make \frac{4x}{25} and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 0.16.
0.16x+0.2y=60,0.16x+0.16y=2.4
Simplify.
0.16x-0.16x+0.2y-0.16y=60-2.4
Subtract 0.16x+0.16y=2.4 from 0.16x+0.2y=60 by subtracting like terms on each side of the equal sign.
0.2y-0.16y=60-2.4
Add \frac{4x}{25} to -\frac{4x}{25}. Terms \frac{4x}{25} and -\frac{4x}{25} cancel out, leaving an equation with only one variable that can be solved.
0.04y=60-2.4
Add \frac{y}{5} to -\frac{4y}{25}.
0.04y=57.6
Add 60 to -2.4.
y=1440
Multiply both sides by 25.
x+1440=15
Substitute 1440 for y in x+y=15. Because the resulting equation contains only one variable, you can solve for x directly.
x=-1425
Subtract 1440 from both sides of the equation.
x=-1425,y=1440
The system is now solved.