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10k+b=700,15k+b=4500
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.
10k+b=700
Choose one of the equations and solve it for k by isolating k on the left hand side of the equal sign.
10k=-b+700
Subtract b from both sides of the equation.
k=\frac{1}{10}\left(-b+700\right)
Divide both sides by 10.
k=-\frac{1}{10}b+70
Multiply \frac{1}{10} times -b+700.
15\left(-\frac{1}{10}b+70\right)+b=4500
Substitute -\frac{b}{10}+70 for k in the other equation, 15k+b=4500.
-\frac{3}{2}b+1050+b=4500
Multiply 15 times -\frac{b}{10}+70.
-\frac{1}{2}b+1050=4500
Add -\frac{3b}{2} to b.
-\frac{1}{2}b=3450
Subtract 1050 from both sides of the equation.
b=-6900
Multiply both sides by -2.
k=-\frac{1}{10}\left(-6900\right)+70
Substitute -6900 for b in k=-\frac{1}{10}b+70. Because the resulting equation contains only one variable, you can solve for k directly.
k=690+70
Multiply -\frac{1}{10} times -6900.
k=760
Add 70 to 690.
k=760,b=-6900
The system is now solved.
10k+b=700,15k+b=4500
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}10&1\\15&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}700\\4500\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}10&1\\15&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}700\\4500\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}10&1\\15&1\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}700\\4500\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}k\\b\end{matrix}\right)=inverse(\left(\begin{matrix}10&1\\15&1\end{matrix}\right))\left(\begin{matrix}700\\4500\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}\frac{1}{10-15}&-\frac{1}{10-15}\\-\frac{15}{10-15}&\frac{10}{10-15}\end{matrix}\right)\left(\begin{matrix}700\\4500\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}k\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}&\frac{1}{5}\\3&-2\end{matrix}\right)\left(\begin{matrix}700\\4500\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{5}\times 700+\frac{1}{5}\times 4500\\3\times 700-2\times 4500\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}k\\b\end{matrix}\right)=\left(\begin{matrix}760\\-6900\end{matrix}\right)
Do the arithmetic.
k=760,b=-6900
Extract the matrix elements k and b.
10k+b=700,15k+b=4500
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.
10k-15k+b-b=700-4500
Subtract 15k+b=4500 from 10k+b=700 by subtracting like terms on each side of the equal sign.
10k-15k=700-4500
Add b to -b. Terms b and -b cancel out, leaving an equation with only one variable that can be solved.
-5k=700-4500
Add 10k to -15k.
-5k=-3800
Add 700 to -4500.
k=760
Divide both sides by -5.
15\times 760+b=4500
Substitute 760 for k in 15k+b=4500. Because the resulting equation contains only one variable, you can solve for b directly.
11400+b=4500
Multiply 15 times 760.
b=-6900
Subtract 11400 from both sides of the equation.
k=760,b=-6900
The system is now solved.