Type a math problem  Solve for a, b  Steps Using Substitution
Steps Using Matrices
Steps Using Elimination
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.
Choose one of the equations and solve it for by isolating on the left hand side of the equal sign.
Subtract from both sides of the equation.
Divide both sides of the equation by , which is the same as multiplying both sides by the reciprocal of the fraction.
Multiply times .
Substitute for in the other equation, .
Subtract from both sides of the equation.
Multiply both sides by .
Substitute for in . Because the resulting equation contains only one variable, you can solve for directly.
Multiply times .
Add to by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
The system is now solved. Giving is as easy as 1, 2, 3
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\frac{3}{2}a+b=1,a+\frac{1}{2}b=7
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.
\frac{3}{2}a+b=1
Choose one of the equations and solve it for a by isolating a on the left hand side of the equal sign.
\frac{3}{2}a=-b+1
Subtract b from both sides of the equation.
a=\frac{2}{3}\left(-b+1\right)
Divide both sides of the equation by \frac{3}{2}=1.5, which is the same as multiplying both sides by the reciprocal of the fraction.
a=-\frac{2}{3}b+\frac{2}{3}
Multiply \frac{2}{3}\approx 0.666666667 times -b+1.
-\frac{2}{3}b+\frac{2}{3}+\frac{1}{2}b=7
Substitute \frac{-2b+2}{3} for a in the other equation, a+\frac{1}{2}b=7.
-\frac{1}{6}b+\frac{2}{3}=7
-\frac{1}{6}b=\frac{19}{3}
Subtract \frac{2}{3}\approx 0.666666667 from both sides of the equation.
b=-38
Multiply both sides by -6.
a=-\frac{2}{3}\left(-38\right)+\frac{2}{3}
Substitute -38 for b in a=-\frac{2}{3}b+\frac{2}{3}. Because the resulting equation contains only one variable, you can solve for a directly.
a=\frac{76+2}{3}
Multiply -\frac{2}{3}\approx -0.666666667 times -38.
a=26
Add \frac{2}{3}\approx 0.666666667 to \frac{76}{3}\approx 25.333333333 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
a=26,b=-38
The system is now solved.
\frac{3}{2}a+b=1,a+\frac{1}{2}b=7
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}1\\7\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right)\left(\begin{matrix}a\\b\end{matrix}\right)=inverse(\left(\begin{matrix}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}1\\7\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}\frac{3}{2}&1\\1&\frac{1}{2}\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}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}1\\7\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}\frac{3}{2}&1\\1&\frac{1}{2}\end{matrix}\right))\left(\begin{matrix}1\\7\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{\frac{1}{2}}{\frac{3}{2}\times \left(\frac{1}{2}\right)-1}&-\frac{1}{\frac{3}{2}\times \left(\frac{1}{2}\right)-1}\\-\frac{1}{\frac{3}{2}\times \left(\frac{1}{2}\right)-1}&\frac{\frac{3}{2}}{\frac{3}{2}\times \left(\frac{1}{2}\right)-1}\end{matrix}\right)\left(\begin{matrix}1\\7\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}-2&4\\4&-6\end{matrix}\right)\left(\begin{matrix}1\\7\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}-2+4\times 7\\4-6\times 7\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a\\b\end{matrix}\right)=\left(\begin{matrix}26\\-38\end{matrix}\right)
Do the arithmetic.
a=26,b=-38
Extract the matrix elements a and b.
\frac{3}{2}a+b=1,a+\frac{1}{2}b=7
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.
\frac{3}{2}a+b=1,\frac{3}{2}a+\frac{3}{2}\times \left(\frac{1}{2}\right)b=\frac{3}{2}\times 7
To make \frac{3a}{2} and a equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by \frac{3}{2}=1.5.
\frac{3}{2}a+b=1,\frac{3}{2}a+\frac{3}{4}b=\frac{21}{2}
Simplify.
\frac{3}{2}a-\frac{3}{2}a+b-\frac{3}{4}b=1-\frac{21}{2}
Subtract \frac{3}{2}a+\frac{3}{4}b=\frac{21}{2} from \frac{3}{2}a+b=1 by subtracting like terms on each side of the equal sign.
b-\frac{3}{4}b=1-\frac{21}{2}
Add \frac{3a}{2} to -\frac{3a}{2}. Terms \frac{3a}{2} and -\frac{3a}{2} cancel out, leaving an equation with only one variable that can be solved.
\frac{1}{4}b=1-\frac{21}{2}