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0.7x+4.5y=75000,x+y=40000
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.
0.7x+4.5y=75000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
0.7x=-4.5y+75000
Subtract \frac{9y}{2} from both sides of the equation.
x=\frac{10}{7}\left(-4.5y+75000\right)
Divide both sides of the equation by 0.7, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{45}{7}y+\frac{750000}{7}
Multiply \frac{10}{7} times -\frac{9y}{2}+75000.
-\frac{45}{7}y+\frac{750000}{7}+y=40000
Substitute \frac{-45y+750000}{7} for x in the other equation, x+y=40000.
-\frac{38}{7}y+\frac{750000}{7}=40000
Add -\frac{45y}{7} to y.
-\frac{38}{7}y=-\frac{470000}{7}
Subtract \frac{750000}{7} from both sides of the equation.
y=\frac{235000}{19}
Divide both sides of the equation by -\frac{38}{7}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{45}{7}\times \frac{235000}{19}+\frac{750000}{7}
Substitute \frac{235000}{19} for y in x=-\frac{45}{7}y+\frac{750000}{7}. Because the resulting equation contains only one variable, you can solve for x directly.
x=-\frac{10575000}{133}+\frac{750000}{7}
Multiply -\frac{45}{7} times \frac{235000}{19} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{525000}{19}
Add \frac{750000}{7} to -\frac{10575000}{133} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{525000}{19},y=\frac{235000}{19}
The system is now solved.
0.7x+4.5y=75000,x+y=40000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}0.7&4.5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}75000\\40000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}0.7&4.5\\1&1\end{matrix}\right))\left(\begin{matrix}0.7&4.5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}0.7&4.5\\1&1\end{matrix}\right))\left(\begin{matrix}75000\\40000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}0.7&4.5\\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}0.7&4.5\\1&1\end{matrix}\right))\left(\begin{matrix}75000\\40000\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}0.7&4.5\\1&1\end{matrix}\right))\left(\begin{matrix}75000\\40000\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}{0.7-4.5}&-\frac{4.5}{0.7-4.5}\\-\frac{1}{0.7-4.5}&\frac{0.7}{0.7-4.5}\end{matrix}\right)\left(\begin{matrix}75000\\40000\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{5}{19}&\frac{45}{38}\\\frac{5}{19}&-\frac{7}{38}\end{matrix}\right)\left(\begin{matrix}75000\\40000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{19}\times 75000+\frac{45}{38}\times 40000\\\frac{5}{19}\times 75000-\frac{7}{38}\times 40000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{525000}{19}\\\frac{235000}{19}\end{matrix}\right)
Do the arithmetic.
x=\frac{525000}{19},y=\frac{235000}{19}
Extract the matrix elements x and y.
0.7x+4.5y=75000,x+y=40000
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.
0.7x+4.5y=75000,0.7x+0.7y=0.7\times 40000
To make \frac{7x}{10} 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 0.7.
0.7x+4.5y=75000,0.7x+0.7y=28000
Simplify.
0.7x-0.7x+4.5y-0.7y=75000-28000
Subtract 0.7x+0.7y=28000 from 0.7x+4.5y=75000 by subtracting like terms on each side of the equal sign.
4.5y-0.7y=75000-28000
Add \frac{7x}{10} to -\frac{7x}{10}. Terms \frac{7x}{10} and -\frac{7x}{10} cancel out, leaving an equation with only one variable that can be solved.
3.8y=75000-28000
Add \frac{9y}{2} to -\frac{7y}{10}.
3.8y=47000
Add 75000 to -28000.
y=\frac{235000}{19}
Divide both sides of the equation by 3.8, which is the same as multiplying both sides by the reciprocal of the fraction.
x+\frac{235000}{19}=40000
Substitute \frac{235000}{19} for y in x+y=40000. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{525000}{19}
Subtract \frac{235000}{19} from both sides of the equation.
x=\frac{525000}{19},y=\frac{235000}{19}
The system is now solved.