Solve for x, y
x=-26750
y=27050
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3x+5y=55000,x+y=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.
3x+5y=55000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
3x=-5y+55000
Subtract 5y from both sides of the equation.
x=\frac{1}{3}\left(-5y+55000\right)
Divide both sides by 3.
x=-\frac{5}{3}y+\frac{55000}{3}
Multiply \frac{1}{3} times -5y+55000.
-\frac{5}{3}y+\frac{55000}{3}+y=300
Substitute \frac{-5y+55000}{3} for x in the other equation, x+y=300.
-\frac{2}{3}y+\frac{55000}{3}=300
Add -\frac{5y}{3} to y.
-\frac{2}{3}y=-\frac{54100}{3}
Subtract \frac{55000}{3} from both sides of the equation.
y=27050
Divide both sides of the equation by -\frac{2}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{5}{3}\times 27050+\frac{55000}{3}
Substitute 27050 for y in x=-\frac{5}{3}y+\frac{55000}{3}. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{-135250+55000}{3}
Multiply -\frac{5}{3} times 27050.
x=-26750
Add \frac{55000}{3} to -\frac{135250}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=-26750,y=27050
The system is now solved.
3x+5y=55000,x+y=300
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}3&5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}55000\\300\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}3&5\\1&1\end{matrix}\right))\left(\begin{matrix}3&5\\1&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}3&5\\1&1\end{matrix}\right))\left(\begin{matrix}55000\\300\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}3&5\\1&1\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}3&5\\1&1\end{matrix}\right))\left(\begin{matrix}55000\\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}3&5\\1&1\end{matrix}\right))\left(\begin{matrix}55000\\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{1}{3-5}&-\frac{5}{3-5}\\-\frac{1}{3-5}&\frac{3}{3-5}\end{matrix}\right)\left(\begin{matrix}55000\\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{1}{2}&\frac{5}{2}\\\frac{1}{2}&-\frac{3}{2}\end{matrix}\right)\left(\begin{matrix}55000\\300\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{2}\times 55000+\frac{5}{2}\times 300\\\frac{1}{2}\times 55000-\frac{3}{2}\times 300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-26750\\27050\end{matrix}\right)
Do the arithmetic.
x=-26750,y=27050
Extract the matrix elements x and y.
3x+5y=55000,x+y=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.
3x+5y=55000,3x+3y=3\times 300
To make 3x and x equal, multiply all terms on each side of the first equation by 1 and all terms on each side of the second by 3.
3x+5y=55000,3x+3y=900
Simplify.
3x-3x+5y-3y=55000-900
Subtract 3x+3y=900 from 3x+5y=55000 by subtracting like terms on each side of the equal sign.
5y-3y=55000-900
Add 3x to -3x. Terms 3x and -3x cancel out, leaving an equation with only one variable that can be solved.
2y=55000-900
Add 5y to -3y.
2y=54100
Add 55000 to -900.
y=27050
Divide both sides by 2.
x+27050=300
Substitute 27050 for y in x+y=300. Because the resulting equation contains only one variable, you can solve for x directly.
x=-26750
Subtract 27050 from both sides of the equation.
x=-26750,y=27050
The system is now solved.
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Matrix
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Simultaneous equation
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Differentiation
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Integration
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Limits
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