Solve for x, y
x = \frac{859}{130} = 6\frac{79}{130} \approx 6.607692308
y = \frac{3769}{130} = 28\frac{129}{130} \approx 28.992307692
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x+y=35.6,0.258x+0.362y=12.2
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=35.6
Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.
x=-y+35.6
Subtract y from both sides of the equation.
0.258\left(-y+35.6\right)+0.362y=12.2
Substitute -y+35.6 for x in the other equation, 0.258x+0.362y=12.2.
-0.258y+9.1848+0.362y=12.2
Multiply 0.258 times -y+35.6.
0.104y+9.1848=12.2
Add -\frac{129y}{500} to \frac{181y}{500}.
0.104y=3.0152
Subtract 9.1848 from both sides of the equation.
y=\frac{3769}{130}
Divide both sides of the equation by 0.104, which is the same as multiplying both sides by the reciprocal of the fraction.
x=-\frac{3769}{130}+35.6
Substitute \frac{3769}{130} for y in x=-y+35.6. Because the resulting equation contains only one variable, you can solve for x directly.
x=\frac{859}{130}
Add 35.6 to -\frac{3769}{130} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
x=\frac{859}{130},y=\frac{3769}{130}
The system is now solved.
x+y=35.6,0.258x+0.362y=12.2
Put the equations in standard form and then use matrices to solve the system of equations.
\left(\begin{matrix}1&1\\0.258&0.362\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}35.6\\12.2\end{matrix}\right)
Write the equations in matrix form.
inverse(\left(\begin{matrix}1&1\\0.258&0.362\end{matrix}\right))\left(\begin{matrix}1&1\\0.258&0.362\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}1&1\\0.258&0.362\end{matrix}\right))\left(\begin{matrix}35.6\\12.2\end{matrix}\right)
Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&1\\0.258&0.362\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\\0.258&0.362\end{matrix}\right))\left(\begin{matrix}35.6\\12.2\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\\0.258&0.362\end{matrix}\right))\left(\begin{matrix}35.6\\12.2\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{0.362}{0.362-0.258}&-\frac{1}{0.362-0.258}\\-\frac{0.258}{0.362-0.258}&\frac{1}{0.362-0.258}\end{matrix}\right)\left(\begin{matrix}35.6\\12.2\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{181}{52}&-\frac{125}{13}\\-\frac{129}{52}&\frac{125}{13}\end{matrix}\right)\left(\begin{matrix}35.6\\12.2\end{matrix}\right)
Do the arithmetic.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{181}{52}\times 35.6-\frac{125}{13}\times 12.2\\-\frac{129}{52}\times 35.6+\frac{125}{13}\times 12.2\end{matrix}\right)
Multiply the matrices.
\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{859}{130}\\\frac{3769}{130}\end{matrix}\right)
Do the arithmetic.
x=\frac{859}{130},y=\frac{3769}{130}
Extract the matrix elements x and y.
x+y=35.6,0.258x+0.362y=12.2
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.
0.258x+0.258y=0.258\times 35.6,0.258x+0.362y=12.2
To make x and \frac{129x}{500} equal, multiply all terms on each side of the first equation by 0.258 and all terms on each side of the second by 1.
0.258x+0.258y=9.1848,0.258x+0.362y=12.2
Simplify.
0.258x-0.258x+0.258y-0.362y=9.1848-12.2
Subtract 0.258x+0.362y=12.2 from 0.258x+0.258y=9.1848 by subtracting like terms on each side of the equal sign.
0.258y-0.362y=9.1848-12.2
Add \frac{129x}{500} to -\frac{129x}{500}. Terms \frac{129x}{500} and -\frac{129x}{500} cancel out, leaving an equation with only one variable that can be solved.
-0.104y=9.1848-12.2
Add \frac{129y}{500} to -\frac{181y}{500}.
-0.104y=-3.0152
Add 9.1848 to -12.2 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
y=\frac{3769}{130}
Divide both sides of the equation by -0.104, which is the same as multiplying both sides by the reciprocal of the fraction.
0.258x+0.362\times \frac{3769}{130}=12.2
Substitute \frac{3769}{130} for y in 0.258x+0.362y=12.2. Because the resulting equation contains only one variable, you can solve for x directly.
0.258x+\frac{682189}{65000}=12.2
Multiply 0.362 times \frac{3769}{130} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
0.258x=\frac{110811}{65000}
Subtract \frac{682189}{65000} from both sides of the equation.
x=\frac{859}{130}
Divide both sides of the equation by 0.258, which is the same as multiplying both sides by the reciprocal of the fraction.
x=\frac{859}{130},y=\frac{3769}{130}
The system is now solved.
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