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x+y=12000;1,05x+1,07y=12240
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=12000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+12000
Subtract y from both sides of the equation.
1,05\left(-y+12000\right)+1,07y=12240
Substitute -y+12000 for x in the other equation, 1,05x+1,07y=12240.
-1,05y+12600+1,07y=12240
Multiply 1,05 times -y+12000.
0,02y+12600=12240
Add -\frac{21y}{20} to \frac{107y}{100}.
0,02y=-360
Subtract 12600 from both sides of the equation.
y=-18000
Multiply both sides by 50.
x=-\left(-18000\right)+12000
Substitute -18000 for y in x=-y+12000. Because the resulting equation contains only one variable, you can solve for x directly.
x=18000+12000
Multiply -1 times -18000.
x=30000
Add 12000 to 18000.
x=30000;y=-18000
The system is now solved.
x+y=12000;1,05x+1,07y=12240
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\1,05&1,07\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12000\\12240\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\1,05&1,07\end{matrix}\right))\left(\begin{matrix}1&1\\1,05&1,07\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\1,05&1,07\end{matrix}\right))\left(\begin{matrix}12000\\12240\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\1,05&1,07\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\\1,05&1,07\end{matrix}\right))\left(\begin{matrix}12000\\12240\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\\1,05&1,07\end{matrix}\right))\left(\begin{matrix}12000\\12240\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{1,07}{1,07-1,05}&-\frac{1}{1,07-1,05}\\-\frac{1,05}{1,07-1,05}&\frac{1}{1,07-1,05}\end{matrix}\right)\left(\begin{matrix}12000\\12240\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}53,5&-50\\-52,5&50\end{matrix}\right)\left(\begin{matrix}12000\\12240\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}53,5\times 12000-50\times 12240\\-52,5\times 12000+50\times 12240\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}30000\\-18000\end{matrix}\right)
Do the arithmetic.
x=30000;y=-18000
Extract the matrix elements x and y.
x+y=12000;1,05x+1,07y=12240
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.
1,05x+1,05y=1,05\times 12000;1,05x+1,07y=12240
To make x and \frac{21x}{20} equal, multiply all terms on each side of the first equation by 1,05 and all terms on each side of the second by 1.
1,05x+1,05y=12600;1,05x+1,07y=12240
Simplify.
1,05x-1,05x+1,05y-1,07y=12600-12240
Subtract 1,05x+1,07y=12240 from 1,05x+1,05y=12600 by subtracting like terms on each side of the equal sign.
1,05y-1,07y=12600-12240
Add \frac{21x}{20} to -\frac{21x}{20}. Terms \frac{21x}{20} and -\frac{21x}{20} cancel out, leaving an equation with only one variable that can be solved.
-0,02y=12600-12240
Add \frac{21y}{20} to -\frac{107y}{100}.
-0,02y=360
Add 12600 to -12240.
y=-18000
Multiply both sides by -50.
1,05x+1,07\left(-18000\right)=12240
Substitute -18000 for y in 1,05x+1,07y=12240. Because the resulting equation contains only one variable, you can solve for x directly.
1,05x-19260=12240
Multiply 1,07 times -18000.
1,05x=31500
Add 19260 to both sides of the equation.
x=30000
Divide both sides of the equation by 1,05, which is the same as multiplying both sides by the reciprocal of the fraction.
x=30000;y=-18000
The system is now solved.