Skip to main content
Solve for a, b
Tick mark Image

Similar Problems from Web Search

Share

5625a+75b=2250,7056a+84b=2250
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.
5625a+75b=2250
Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.
5625a=-75b+2250
Subtract 75b from both sides of the equation.
a=\frac{1}{5625}\left(-75b+2250\right)
Divide both sides by 5625.
a=-\frac{1}{75}b+\frac{2}{5}
Multiply \frac{1}{5625} times -75b+2250.
7056\left(-\frac{1}{75}b+\frac{2}{5}\right)+84b=2250
Substitute -\frac{b}{75}+\frac{2}{5} for a in the other equation, 7056a+84b=2250.
-\frac{2352}{25}b+\frac{14112}{5}+84b=2250
Multiply 7056 times -\frac{b}{75}+\frac{2}{5}.
-\frac{252}{25}b+\frac{14112}{5}=2250
Add -\frac{2352b}{25} to 84b.
-\frac{252}{25}b=-\frac{2862}{5}
Subtract \frac{14112}{5} from both sides of the equation.
b=\frac{795}{14}
Divide both sides of the equation by -\frac{252}{25}, which is the same as multiplying both sides by the reciprocal of the fraction.
a=-\frac{1}{75}\times \frac{795}{14}+\frac{2}{5}
Substitute \frac{795}{14} for b in a=-\frac{1}{75}b+\frac{2}{5}. Because the resulting equation contains only one variable, you can solve for a directly.
a=-\frac{53}{70}+\frac{2}{5}
Multiply -\frac{1}{75} times \frac{795}{14} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
a=-\frac{5}{14}
Add \frac{2}{5} to -\frac{53}{70} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
a=-\frac{5}{14},b=\frac{795}{14}
The system is now solved.
5625a+75b=2250,7056a+84b=2250
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}2250\\2250\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right))\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right))\left(\begin{matrix}2250\\2250\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}5625&75\\7056&84\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right))\left(\begin{matrix}2250\\2250\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}5625&75\\7056&84\end{matrix}\right))\left(\begin{matrix}2250\\2250\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}\frac{84}{5625\times 84-75\times 7056}&-\frac{75}{5625\times 84-75\times 7056}\\-\frac{7056}{5625\times 84-75\times 7056}&\frac{5625}{5625\times 84-75\times 7056}\end{matrix}\right)\left(\begin{matrix}2250\\2250\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}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{675}&\frac{1}{756}\\\frac{28}{225}&-\frac{25}{252}\end{matrix}\right)\left(\begin{matrix}2250\\2250\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{675}\times 2250+\frac{1}{756}\times 2250\\\frac{28}{225}\times 2250-\frac{25}{252}\times 2250\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{5}{14}\\\frac{795}{14}\end{matrix}\right)
Do the arithmetic.
a=-\frac{5}{14},b=\frac{795}{14}
Extract the matrix elements a and b.
5625a+75b=2250,7056a+84b=2250
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.
7056\times 5625a+7056\times 75b=7056\times 2250,5625\times 7056a+5625\times 84b=5625\times 2250
To make 5625a and 7056a equal, multiply all terms on each side of the first equation by 7056 and all terms on each side of the second by 5625.
39690000a+529200b=15876000,39690000a+472500b=12656250
Simplify.
39690000a-39690000a+529200b-472500b=15876000-12656250
Subtract 39690000a+472500b=12656250 from 39690000a+529200b=15876000 by subtracting like terms on each side of the equal sign.
529200b-472500b=15876000-12656250
Add 39690000a to -39690000a. Terms 39690000a and -39690000a cancel out, leaving an equation with only one variable that can be solved.
56700b=15876000-12656250
Add 529200b to -472500b.
56700b=3219750
Add 15876000 to -12656250.
b=\frac{795}{14}
Divide both sides by 56700.
7056a+84\times \frac{795}{14}=2250
Substitute \frac{795}{14} for b in 7056a+84b=2250. Because the resulting equation contains only one variable, you can solve for a directly.
7056a+4770=2250
Multiply 84 times \frac{795}{14}.
7056a=-2520
Subtract 4770 from both sides of the equation.
a=-\frac{5}{14}
Divide both sides by 7056.
a=-\frac{5}{14},b=\frac{795}{14}
The system is now solved.