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10a_{1}+45d=100,100a_{1}+4950d=10
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.
10a_{1}+45d=100
Choose one of the equations and solve it for a_{1} by isolating a_{1} on the left hand side of the equal sign.
10a_{1}=-45d+100
Subtract 45d from both sides of the equation.
a_{1}=\frac{1}{10}\left(-45d+100\right)
Divide both sides by 10.
a_{1}=-\frac{9}{2}d+10
Multiply \frac{1}{10} times -45d+100.
100\left(-\frac{9}{2}d+10\right)+4950d=10
Substitute -\frac{9d}{2}+10 for a_{1} in the other equation, 100a_{1}+4950d=10.
-450d+1000+4950d=10
Multiply 100 times -\frac{9d}{2}+10.
4500d+1000=10
Add -450d to 4950d.
4500d=-990
Subtract 1000 from both sides of the equation.
d=-\frac{11}{50}
Divide both sides by 4500.
a_{1}=-\frac{9}{2}\left(-\frac{11}{50}\right)+10
Substitute -\frac{11}{50} for d in a_{1}=-\frac{9}{2}d+10. Because the resulting equation contains only one variable, you can solve for a_{1} directly.
a_{1}=\frac{99}{100}+10
Multiply -\frac{9}{2} times -\frac{11}{50} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
a_{1}=\frac{1099}{100}
Add 10 to \frac{99}{100}.
a_{1}=\frac{1099}{100},d=-\frac{11}{50}
The system is now solved.
10a_{1}+45d=100,100a_{1}+4950d=10
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}10&45\\100&4950\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}100\\10\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}10&45\\100&4950\end{matrix}\right))\left(\begin{matrix}10&45\\100&4950\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}10&45\\100&4950\end{matrix}\right))\left(\begin{matrix}100\\10\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&45\\100&4950\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}10&45\\100&4950\end{matrix}\right))\left(\begin{matrix}100\\10\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=inverse(\left(\begin{matrix}10&45\\100&4950\end{matrix}\right))\left(\begin{matrix}100\\10\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{4950}{10\times 4950-45\times 100}&-\frac{45}{10\times 4950-45\times 100}\\-\frac{100}{10\times 4950-45\times 100}&\frac{10}{10\times 4950-45\times 100}\end{matrix}\right)\left(\begin{matrix}100\\10\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_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{11}{100}&-\frac{1}{1000}\\-\frac{1}{450}&\frac{1}{4500}\end{matrix}\right)\left(\begin{matrix}100\\10\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{11}{100}\times 100-\frac{1}{1000}\times 10\\-\frac{1}{450}\times 100+\frac{1}{4500}\times 10\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}a_{1}\\d\end{matrix}\right)=\left(\begin{matrix}\frac{1099}{100}\\-\frac{11}{50}\end{matrix}\right)
Do the arithmetic.
a_{1}=\frac{1099}{100},d=-\frac{11}{50}
Extract the matrix elements a_{1} and d.
10a_{1}+45d=100,100a_{1}+4950d=10
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.
100\times 10a_{1}+100\times 45d=100\times 100,10\times 100a_{1}+10\times 4950d=10\times 10
To make 10a_{1} and 100a_{1} equal, multiply all terms on each side of the first equation by 100 and all terms on each side of the second by 10.
1000a_{1}+4500d=10000,1000a_{1}+49500d=100
Simplify.
1000a_{1}-1000a_{1}+4500d-49500d=10000-100
Subtract 1000a_{1}+49500d=100 from 1000a_{1}+4500d=10000 by subtracting like terms on each side of the equal sign.
4500d-49500d=10000-100
Add 1000a_{1} to -1000a_{1}. Terms 1000a_{1} and -1000a_{1} cancel out, leaving an equation with only one variable that can be solved.
-45000d=10000-100
Add 4500d to -49500d.
-45000d=9900
Add 10000 to -100.
d=-\frac{11}{50}
Divide both sides by -45000.
100a_{1}+4950\left(-\frac{11}{50}\right)=10
Substitute -\frac{11}{50} for d in 100a_{1}+4950d=10. Because the resulting equation contains only one variable, you can solve for a_{1} directly.
100a_{1}-1089=10
Multiply 4950 times -\frac{11}{50}.
100a_{1}=1099
Add 1089 to both sides of the equation.
a_{1}=\frac{1099}{100}
Divide both sides by 100.
a_{1}=\frac{1099}{100},d=-\frac{11}{50}
The system is now solved.