\left\{ \begin{array} { l } { x + y = 100 } \\ { 18 x + 10 y = 75 ( x + 4 ) } \end{array} \right.
Solve for x, y
x = \frac{700}{67} = 10\frac{30}{67} \approx 10.447761194
y = \frac{6000}{67} = 89\frac{37}{67} \approx 89.552238806
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18x+10y=75x+300
Consider the second equation. Use the distributive property to multiply 75 by x+4.
18x+10y-75x=300
Subtract 75x from both sides.
-57x+10y=300
Combine 18x and -75x to get -57x.
x+y=100,-57x+10y=300
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.
x+y=100
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+100
Subtract y from both sides of the equation.
-57\left(-y+100\right)+10y=300
Substitute -y+100 for x in the other equation, -57x+10y=300.
57y-5700+10y=300
Multiply -57 times -y+100.
67y-5700=300
Add 57y to 10y.
67y=6000
Add 5700 to both sides of the equation.
y=\frac{6000}{67}
Divide both sides by 67.
x=-\frac{6000}{67}+100
Substitute \frac{6000}{67} for y in x=-y+100. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{700}{67}
Add 100 to -\frac{6000}{67}.
x=\frac{700}{67},y=\frac{6000}{67}
The system is now solved.
18x+10y=75x+300
Consider the second equation. Use the distributive property to multiply 75 by x+4.
18x+10y-75x=300
Subtract 75x from both sides.
-57x+10y=300
Combine 18x and -75x to get -57x.
x+y=100,-57x+10y=300
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\-57&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}100\\300\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}1&1\\-57&10\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\-57&10\end{matrix}\right).
\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)
The product of a matrix and its inverse is the identity matrix.
\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\-57&10\end{matrix}\right))\left(\begin{matrix}100\\300\end{matrix}\right)
Multiply the matrices on the left hand side of the equal sign.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{10-\left(-57\right)}&-\frac{1}{10-\left(-57\right)}\\-\frac{-57}{10-\left(-57\right)}&\frac{1}{10-\left(-57\right)}\end{matrix}\right)\left(\begin{matrix}100\\300\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}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{67}&-\frac{1}{67}\\\frac{57}{67}&\frac{1}{67}\end{matrix}\right)\left(\begin{matrix}100\\300\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{10}{67}\times 100-\frac{1}{67}\times 300\\\frac{57}{67}\times 100+\frac{1}{67}\times 300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{700}{67}\\\frac{6000}{67}\end{matrix}\right)
Do the arithmetic.
x=\frac{700}{67},y=\frac{6000}{67}
Extract the matrix elements x and y.
18x+10y=75x+300
Consider the second equation. Use the distributive property to multiply 75 by x+4.
18x+10y-75x=300
Subtract 75x from both sides.
-57x+10y=300
Combine 18x and -75x to get -57x.
x+y=100,-57x+10y=300
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.
-57x-57y=-57\times 100,-57x+10y=300
To make x and -57x equal, multiply all terms on each side of the first equation by -57 and all terms on each side of the second by 1.
-57x-57y=-5700,-57x+10y=300
Simplify.
-57x+57x-57y-10y=-5700-300
Subtract -57x+10y=300 from -57x-57y=-5700 by subtracting like terms on each side of the equal sign.
-57y-10y=-5700-300
Add -57x to 57x. Terms -57x and 57x cancel out, leaving an equation with only one variable that can be solved.
-67y=-5700-300
Add -57y to -10y.
-67y=-6000
Add -5700 to -300.
y=\frac{6000}{67}
Divide both sides by -67.
-57x+10\times \frac{6000}{67}=300
Substitute \frac{6000}{67} for y in -57x+10y=300. Because the resulting equation contains only one variable, you can solve for x directly.
-57x+\frac{60000}{67}=300
Multiply 10 times \frac{6000}{67}.
-57x=-\frac{39900}{67}
Subtract \frac{60000}{67} from both sides of the equation.
x=\frac{700}{67}
Divide both sides by -57.
x=\frac{700}{67},y=\frac{6000}{67}
The system is now solved.
Examples
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Matrix
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Simultaneous equation
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Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
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Limits
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