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x-0.05y=170000
Consider the first equation. Subtract 0.05y from both sides.
y-0.1935x=340000
Consider the second equation. Subtract 0.1935x from both sides.
x-0.05y=170000,-0.1935x+y=340000
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-0.05y=170000
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=0.05y+170000
Add \frac{y}{20} to both sides of the equation.
-0.1935\left(0.05y+170000\right)+y=340000
Substitute \frac{y}{20}+170000 for x in the other equation, -0.1935x+y=340000.
-0.009675y-32895+y=340000
Multiply -0.1935 times \frac{y}{20}+170000.
0.990325y-32895=340000
Add -\frac{387y}{40000} to y.
0.990325y=372895
Add 32895 to both sides of the equation.
y=\frac{14915800000}{39613}
Divide both sides of the equation by 0.990325, which is the same as multiplying both sides by the reciprocal of the fraction.
x=0.05\times \frac{14915800000}{39613}+170000
Substitute \frac{14915800000}{39613} for y in x=0.05y+170000. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{745790000}{39613}+170000
Multiply 0.05 times \frac{14915800000}{39613} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
x=\frac{7480000000}{39613}
Add 170000 to \frac{745790000}{39613}.
x=\frac{7480000000}{39613},y=\frac{14915800000}{39613}
The system is now solved.
x-0.05y=170000
Consider the first equation. Subtract 0.05y from both sides.
y-0.1935x=340000
Consider the second equation. Subtract 0.1935x from both sides.
x-0.05y=170000,-0.1935x+y=340000
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&-0.05\\-0.1935&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}170000\\340000\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&-0.05\\-0.1935&1\end{matrix}\right))\left(\begin{matrix}1&-0.05\\-0.1935&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&-0.05\\-0.1935&1\end{matrix}\right))\left(\begin{matrix}170000\\340000\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-0.05\\-0.1935&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}1&-0.05\\-0.1935&1\end{matrix}\right))\left(\begin{matrix}170000\\340000\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&-0.05\\-0.1935&1\end{matrix}\right))\left(\begin{matrix}170000\\340000\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}{1-\left(-0.05\left(-0.1935\right)\right)}&-\frac{-0.05}{1-\left(-0.05\left(-0.1935\right)\right)}\\-\frac{-0.1935}{1-\left(-0.05\left(-0.1935\right)\right)}&\frac{1}{1-\left(-0.05\left(-0.1935\right)\right)}\end{matrix}\right)\left(\begin{matrix}170000\\340000\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{40000}{39613}&\frac{2000}{39613}\\\frac{7740}{39613}&\frac{40000}{39613}\end{matrix}\right)\left(\begin{matrix}170000\\340000\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{40000}{39613}\times 170000+\frac{2000}{39613}\times 340000\\\frac{7740}{39613}\times 170000+\frac{40000}{39613}\times 340000\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{7480000000}{39613}\\\frac{14915800000}{39613}\end{matrix}\right)
Do the arithmetic.
x=\frac{7480000000}{39613},y=\frac{14915800000}{39613}
Extract the matrix elements x and y.
x-0.05y=170000
Consider the first equation. Subtract 0.05y from both sides.
y-0.1935x=340000
Consider the second equation. Subtract 0.1935x from both sides.
x-0.05y=170000,-0.1935x+y=340000
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.
-0.1935x-0.1935\left(-0.05\right)y=-0.1935\times 170000,-0.1935x+y=340000
To make x and -\frac{387x}{2000} equal, multiply all terms on each side of the first equation by -0.1935 and all terms on each side of the second by 1.
-0.1935x+0.009675y=-32895,-0.1935x+y=340000
Simplify.
-0.1935x+0.1935x+0.009675y-y=-32895-340000
Subtract -0.1935x+y=340000 from -0.1935x+0.009675y=-32895 by subtracting like terms on each side of the equal sign.
0.009675y-y=-32895-340000
Add -\frac{387x}{2000} to \frac{387x}{2000}. Terms -\frac{387x}{2000} and \frac{387x}{2000} cancel out, leaving an equation with only one variable that can be solved.
-0.990325y=-32895-340000
Add \frac{387y}{40000} to -y.
-0.990325y=-372895
Add -32895 to -340000.
y=\frac{14915800000}{39613}
Divide both sides of the equation by -0.990325, which is the same as multiplying both sides by the reciprocal of the fraction.
-0.1935x+\frac{14915800000}{39613}=340000
Substitute \frac{14915800000}{39613} for y in -0.1935x+y=340000. Because the resulting equation contains only one variable, you can solve for x directly.
-0.1935x=-\frac{1447380000}{39613}
Subtract \frac{14915800000}{39613} from both sides of the equation.
x=\frac{7480000000}{39613}
Divide both sides of the equation by -0.1935, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{7480000000}{39613},y=\frac{14915800000}{39613}
The system is now solved.