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50\times \frac{1}{1000}\times 2\times 3000=x+y
Consider the first equation. Calculate 10 to the power of -3 and get \frac{1}{1000}.
\frac{1}{20}\times 2\times 3000=x+y
Multiply 50 and \frac{1}{1000} to get \frac{1}{20}.
\frac{1}{10}\times 3000=x+y
Multiply \frac{1}{20} and 2 to get \frac{1}{10}.
300=x+y
Multiply \frac{1}{10} and 3000 to get 300.
x+y=300
Swap sides so that all variable terms are on the left hand side.
0.4x-0.6y=0
Consider the second equation. Subtract 0.6y from both sides.
x+y=300,0.4x-0.6y=0
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=300
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+300
Subtract y from both sides of the equation.
0.4\left(-y+300\right)-0.6y=0
Substitute -y+300 for x in the other equation, 0.4x-0.6y=0.
-0.4y+120-0.6y=0
Multiply 0.4 times -y+300.
-y+120=0
Add -\frac{2y}{5} to -\frac{3y}{5}.
-y=-120
Subtract 120 from both sides of the equation.
y=120
Divide both sides by -1.
x=-120+300
Substitute 120 for y in x=-y+300. Because the resulting equation contains only one variable, you can solve for x directly.
x=180
Add 300 to -120.
x=180,y=120
The system is now solved.
50\times \frac{1}{1000}\times 2\times 3000=x+y
Consider the first equation. Calculate 10 to the power of -3 and get \frac{1}{1000}.
\frac{1}{20}\times 2\times 3000=x+y
Multiply 50 and \frac{1}{1000} to get \frac{1}{20}.
\frac{1}{10}\times 3000=x+y
Multiply \frac{1}{20} and 2 to get \frac{1}{10}.
300=x+y
Multiply \frac{1}{10} and 3000 to get 300.
x+y=300
Swap sides so that all variable terms are on the left hand side.
0.4x-0.6y=0
Consider the second equation. Subtract 0.6y from both sides.
x+y=300,0.4x-0.6y=0
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\0.4&-0.6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}300\\0\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\0.4&-0.6\end{matrix}\right))\left(\begin{matrix}1&1\\0.4&-0.6\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\0.4&-0.6\end{matrix}\right))\left(\begin{matrix}300\\0\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\0.4&-0.6\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\\0.4&-0.6\end{matrix}\right))\left(\begin{matrix}300\\0\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\\0.4&-0.6\end{matrix}\right))\left(\begin{matrix}300\\0\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{0.6}{-0.6-0.4}&-\frac{1}{-0.6-0.4}\\-\frac{0.4}{-0.6-0.4}&\frac{1}{-0.6-0.4}\end{matrix}\right)\left(\begin{matrix}300\\0\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}0.6&1\\0.4&-1\end{matrix}\right)\left(\begin{matrix}300\\0\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0.6\times 300\\0.4\times 300\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}180\\120\end{matrix}\right)
Do the arithmetic.
x=180,y=120
Extract the matrix elements x and y.
50\times \frac{1}{1000}\times 2\times 3000=x+y
Consider the first equation. Calculate 10 to the power of -3 and get \frac{1}{1000}.
\frac{1}{20}\times 2\times 3000=x+y
Multiply 50 and \frac{1}{1000} to get \frac{1}{20}.
\frac{1}{10}\times 3000=x+y
Multiply \frac{1}{20} and 2 to get \frac{1}{10}.
300=x+y
Multiply \frac{1}{10} and 3000 to get 300.
x+y=300
Swap sides so that all variable terms are on the left hand side.
0.4x-0.6y=0
Consider the second equation. Subtract 0.6y from both sides.
x+y=300,0.4x-0.6y=0
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.
0.4x+0.4y=0.4\times 300,0.4x-0.6y=0
To make x and \frac{2x}{5} equal, multiply all terms on each side of the first equation by 0.4 and all terms on each side of the second by 1.
0.4x+0.4y=120,0.4x-0.6y=0
Simplify.
0.4x-0.4x+0.4y+0.6y=120
Subtract 0.4x-0.6y=0 from 0.4x+0.4y=120 by subtracting like terms on each side of the equal sign.
0.4y+0.6y=120
Add \frac{2x}{5} to -\frac{2x}{5}. Terms \frac{2x}{5} and -\frac{2x}{5} cancel out, leaving an equation with only one variable that can be solved.
y=120
Add \frac{2y}{5} to \frac{3y}{5}.
0.4x-0.6\times 120=0
Substitute 120 for y in 0.4x-0.6y=0. Because the resulting equation contains only one variable, you can solve for x directly.
0.4x-72=0
Multiply -0.6 times 120.
0.4x=72
Add 72 to both sides of the equation.
x=180
Divide both sides of the equation by 0.4, which is the same as multiplying both sides by the reciprocal of the fraction.
x=180,y=120
The system is now solved.