\left\{ \begin{array} { l } { x - y = 12000 } \\ { 1,2 x - 0,9 y = 23400 } \end{array} \right.
Solve for x, y
x=42000
y=30000
Graph
Share
Copied to clipboard
x-y=12000;1,2x-0,9y=23400
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
Add y to both sides of the equation.
1,2\left(y+12000\right)-0,9y=23400
Substitute y+12000 for x in the other equation, 1,2x-0,9y=23400.
1,2y+14400-0,9y=23400
Multiply 1,2 times y+12000.
0,3y+14400=23400
Add \frac{6y}{5} to -\frac{9y}{10}.
0,3y=9000
Subtract 14400 from both sides of the equation.
y=30000
Divide both sides of the equation by 0,3, which is the same as multiplying both sides by the reciprocal of the fraction.
x=30000+12000
Substitute 30000 for y in x=y+12000. Because the resulting equation contains only one variable, you can solve for x directly.
x=42000
Add 12000 to 30000.
x=42000;y=30000
The system is now solved.
x-y=12000;1,2x-0,9y=23400
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-1\\1,2&-0,9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}12000\\23400\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-1\\1,2&-0,9\end{matrix}\right))\left(\begin{matrix}1&-1\\1,2&-0,9\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-1\\1,2&-0,9\end{matrix}\right))\left(\begin{matrix}12000\\23400\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-1\\1,2&-0,9\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\\1,2&-0,9\end{matrix}\right))\left(\begin{matrix}12000\\23400\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\\1,2&-0,9\end{matrix}\right))\left(\begin{matrix}12000\\23400\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{0,9}{-0,9-\left(-1,2\right)}&-\frac{-1}{-0,9-\left(-1,2\right)}\\-\frac{1,2}{-0,9-\left(-1,2\right)}&\frac{1}{-0,9-\left(-1,2\right)}\end{matrix}\right)\left(\begin{matrix}12000\\23400\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&\frac{10}{3}\\-4&\frac{10}{3}\end{matrix}\right)\left(\begin{matrix}12000\\23400\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-3\times 12000+\frac{10}{3}\times 23400\\-4\times 12000+\frac{10}{3}\times 23400\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}42000\\30000\end{matrix}\right)
Do the arithmetic.
x=42000;y=30000
Extract the matrix elements x and y.
x-y=12000;1,2x-0,9y=23400
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.
1,2x+1,2\left(-1\right)y=1,2\times 12000;1,2x-0,9y=23400
To make x and \frac{6x}{5} equal, multiply all terms on each side of the first equation by 1,2 and all terms on each side of the second by 1.
1,2x-1,2y=14400;1,2x-0,9y=23400
Simplify.
1,2x-1,2x-1,2y+0,9y=14400-23400
Subtract 1,2x-0,9y=23400 from 1,2x-1,2y=14400 by subtracting like terms on each side of the equal sign.
-1,2y+0,9y=14400-23400
Add \frac{6x}{5} to -\frac{6x}{5}. Terms \frac{6x}{5} and -\frac{6x}{5} cancel out, leaving an equation with only one variable that can be solved.
-0,3y=14400-23400
Add -\frac{6y}{5} to \frac{9y}{10}.
-0,3y=-9000
Add 14400 to -23400.
y=30000
Divide both sides of the equation by -0,3, which is the same as multiplying both sides by the reciprocal of the fraction.
1,2x-0,9\times 30000=23400
Substitute 30000 for y in 1,2x-0,9y=23400. Because the resulting equation contains only one variable, you can solve for x directly.
1,2x-27000=23400
Multiply -0,9 times 30000.
1,2x=50400
Add 27000 to both sides of the equation.
x=42000
Divide both sides of the equation by 1,2, which is the same as multiplying both sides by the reciprocal of the fraction.
x=42000;y=30000
The system is now solved.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}