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4b-2x=126,2b+2x=84
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.
4b-2x=126
Choose one of the equations and solve it for b by isolating b on the left hand side of the equal sign.
4b=2x+126
Add 2x to both sides of the equation.
b=\frac{1}{4}\left(2x+126\right)
Divide both sides by 4.
b=\frac{1}{2}x+\frac{63}{2}
Multiply \frac{1}{4} times 126+2x.
2\left(\frac{1}{2}x+\frac{63}{2}\right)+2x=84
Substitute \frac{63+x}{2} for b in the other equation, 2b+2x=84.
x+63+2x=84
Multiply 2 times \frac{63+x}{2}.
3x+63=84
Add x to 2x.
3x=21
Subtract 63 from both sides of the equation.
x=7
Divide both sides by 3.
b=\frac{1}{2}\times 7+\frac{63}{2}
Substitute 7 for x in b=\frac{1}{2}x+\frac{63}{2}. Because the resulting equation contains only one variable, you can solve for b directly.
b=\frac{7+63}{2}
Multiply \frac{1}{2} times 7.
b=35
Add \frac{63}{2} to \frac{7}{2} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
b=35,x=7
The system is now solved.
4b-2x=126,2b+2x=84
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}4&-2\\2&2\end{matrix}\right)\left(\begin{matrix}b\\x\end{matrix}\right)=\left(\begin{matrix}126\\84\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}4&-2\\2&2\end{matrix}\right))\left(\begin{matrix}4&-2\\2&2\end{matrix}\right)\left(\begin{matrix}b\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\2&2\end{matrix}\right))\left(\begin{matrix}126\\84\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}4&-2\\2&2\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}b\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\2&2\end{matrix}\right))\left(\begin{matrix}126\\84\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}b\\x\end{matrix}\right)=inverse(\left(\begin{matrix}4&-2\\2&2\end{matrix}\right))\left(\begin{matrix}126\\84\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}b\\x\end{matrix}\right)=\left(\begin{matrix}\frac{2}{4\times 2-\left(-2\times 2\right)}&-\frac{-2}{4\times 2-\left(-2\times 2\right)}\\-\frac{2}{4\times 2-\left(-2\times 2\right)}&\frac{4}{4\times 2-\left(-2\times 2\right)}\end{matrix}\right)\left(\begin{matrix}126\\84\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}b\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}&\frac{1}{6}\\-\frac{1}{6}&\frac{1}{3}\end{matrix}\right)\left(\begin{matrix}126\\84\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}b\\x\end{matrix}\right)=\left(\begin{matrix}\frac{1}{6}\times 126+\frac{1}{6}\times 84\\-\frac{1}{6}\times 126+\frac{1}{3}\times 84\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}b\\x\end{matrix}\right)=\left(\begin{matrix}35\\7\end{matrix}\right)
Do the arithmetic.
b=35,x=7
Extract the matrix elements b and x.
4b-2x=126,2b+2x=84
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\times 4b+2\left(-2\right)x=2\times 126,4\times 2b+4\times 2x=4\times 84
To make 4b and 2b equal, multiply all terms on each side of the first equation by 2 and all terms on each side of the second by 4.
8b-4x=252,8b+8x=336
Simplify.
8b-8b-4x-8x=252-336
Subtract 8b+8x=336 from 8b-4x=252 by subtracting like terms on each side of the equal sign.
-4x-8x=252-336
Add 8b to -8b. Terms 8b and -8b cancel out, leaving an equation with only one variable that can be solved.
-12x=252-336
Add -4x to -8x.
-12x=-84
Add 252 to -336.
x=7
Divide both sides by -12.
2b+2\times 7=84
Substitute 7 for x in 2b+2x=84. Because the resulting equation contains only one variable, you can solve for b directly.
2b+14=84
Multiply 2 times 7.
2b=70
Subtract 14 from both sides of the equation.
b=35
Divide both sides by 2.
b=35,x=7
The system is now solved.