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x+y=2000,4x+3y=4600
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=2000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+2000
Subtract y from both sides of the equation.
4\left(-y+2000\right)+3y=4600
Substitute -y+2000 for x in the other equation, 4x+3y=4600.
-4y+8000+3y=4600
Multiply 4 times -y+2000.
-y+8000=4600
Add -4y to 3y.
-y=-3400
Subtract 8000 from both sides of the equation.
y=3400
Divide both sides by -1.
x=-3400+2000
Substitute 3400 for y in x=-y+2000. Because the resulting equation contains only one variable, you can solve for x directly.
x=-1400
Add 2000 to -3400.
x=-1400,y=3400
The system is now solved.
x+y=2000,4x+3y=4600
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}2000\\4600\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}1&1\\4&3\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\4&3\end{matrix}\right))\left(\begin{matrix}2000\\4600\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\4&3\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\\4&3\end{matrix}\right))\left(\begin{matrix}2000\\4600\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\\4&3\end{matrix}\right))\left(\begin{matrix}2000\\4600\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{3}{3-4}&-\frac{1}{3-4}\\-\frac{4}{3-4}&\frac{1}{3-4}\end{matrix}\right)\left(\begin{matrix}2000\\4600\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}-3&1\\4&-1\end{matrix}\right)\left(\begin{matrix}2000\\4600\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 2000+4600\\4\times 2000-4600\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-1400\\3400\end{matrix}\right)
Do the arithmetic.
x=-1400,y=3400
Extract the matrix elements x and y.
x+y=2000,4x+3y=4600
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.
4x+4y=4\times 2000,4x+3y=4600
To make x and 4x equal, multiply all terms on each side of the first equation by 4 and all terms on each side of the second by 1.
4x+4y=8000,4x+3y=4600
Simplify.
4x-4x+4y-3y=8000-4600
Subtract 4x+3y=4600 from 4x+4y=8000 by subtracting like terms on each side of the equal sign.
4y-3y=8000-4600
Add 4x to -4x. Terms 4x and -4x cancel out, leaving an equation with only one variable that can be solved.
y=8000-4600
Add 4y to -3y.
y=3400
Add 8000 to -4600.
4x+3\times 3400=4600
Substitute 3400 for y in 4x+3y=4600. Because the resulting equation contains only one variable, you can solve for x directly.
4x+10200=4600
Multiply 3 times 3400.
4x=-5600
Subtract 10200 from both sides of the equation.
x=-1400
Divide both sides by 4.
x=-1400,y=3400
The system is now solved.