Solve {l}{2x-3y=68}{26x+5y=1}

$$\left. \begin{array} { l } { 2 x - 3 y = 68 } \\ { 26 x + 5 y = 1 } \end{array} \right.$$

Solve for x, y

$$x = \frac{343}{88} = 3\frac{79}{88} \approx 3.897727273$$

$$y = -\frac{883}{44} = -20\frac{3}{44} \approx -20.068181818$$

Steps Using Substitution

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.

$$2x-3y=68,26x+5y=1$$

Choose one of the equations and solve it for x by isolating x on the left hand side of the equal sign.

$$2x-3y=68$$

Add 3y to both sides of the equation.

$$2x=3y+68$$

Divide both sides by 2.

$$x=\frac{1}{2}\left(3y+68\right)$$

Multiply \frac{1}{2} times 3y+68.

$$x=\frac{3}{2}y+34$$

Substitute \frac{3y}{2}+34 for x in the other equation, 26x+5y=1.

$$26\left(\frac{3}{2}y+34\right)+5y=1$$

Multiply 26 times \frac{3y}{2}+34.

$$39y+884+5y=1$$

Add 39y to 5y.

$$44y+884=1$$

Subtract 884 from both sides of the equation.

$$44y=-883$$

Divide both sides by 44.

$$y=-\frac{883}{44}$$

Substitute -\frac{883}{44} for y in x=\frac{3}{2}y+34. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=\frac{3}{2}\left(-\frac{883}{44}\right)+34$$

Multiply \frac{3}{2} times -\frac{883}{44} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{2649}{88}+34$$

Add 34 to -\frac{2649}{88}.

$$x=\frac{343}{88}$$

The system is now solved.

$$x=\frac{343}{88},y=-\frac{883}{44}$$

Steps Using Matrices

Put the equations in standard form and then use matrices to solve the system of equations.

$$2x-3y=68,26x+5y=1$$

Write the equations in matrix form.

$$\left(\begin{matrix}2&-3\\26&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}68\\1\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}2&-3\\26&5\end{matrix}\right).

$$inverse(\left(\begin{matrix}2&-3\\26&5\end{matrix}\right))\left(\begin{matrix}2&-3\\26&5\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\26&5\end{matrix}\right))\left(\begin{matrix}68\\1\end{matrix}\right)$$

The product of a matrix and its inverse is the identity matrix.

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\26&5\end{matrix}\right))\left(\begin{matrix}68\\1\end{matrix}\right)$$

Multiply the matrices on the left hand side of the equal sign.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}2&-3\\26&5\end{matrix}\right))\left(\begin{matrix}68\\1\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{2\times 5-\left(-3\times 26\right)}&-\frac{-3}{2\times 5-\left(-3\times 26\right)}\\-\frac{26}{2\times 5-\left(-3\times 26\right)}&\frac{2}{2\times 5-\left(-3\times 26\right)}\end{matrix}\right)\left(\begin{matrix}68\\1\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{88}&\frac{3}{88}\\-\frac{13}{44}&\frac{1}{44}\end{matrix}\right)\left(\begin{matrix}68\\1\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{5}{88}\times 68+\frac{3}{88}\\-\frac{13}{44}\times 68+\frac{1}{44}\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{343}{88}\\-\frac{883}{44}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=\frac{343}{88},y=-\frac{883}{44}$$

Steps Using Elimination

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.

$$2x-3y=68,26x+5y=1$$

To make 2x and 26x equal, multiply all terms on each side of the first equation by 26 and all terms on each side of the second by 2.

$$26\times 2x+26\left(-3\right)y=26\times 68,2\times 26x+2\times 5y=2$$

Simplify.

$$52x-78y=1768,52x+10y=2$$

Subtract 52x+10y=2 from 52x-78y=1768 by subtracting like terms on each side of the equal sign.

$$52x-52x-78y-10y=1768-2$$

Add 52x to -52x. Terms 52x and -52x cancel out, leaving an equation with only one variable that can be solved.

$$-78y-10y=1768-2$$

Add -78y to -10y.

$$-88y=1768-2$$

Add 1768 to -2.

$$-88y=1766$$

Divide both sides by -88.

$$y=-\frac{883}{44}$$

Substitute -\frac{883}{44} for y in 26x+5y=1. Because the resulting equation contains only one variable, you can solve for x directly.

$$26x+5\left(-\frac{883}{44}\right)=1$$

Multiply 5 times -\frac{883}{44}.

$$26x-\frac{4415}{44}=1$$

Add \frac{4415}{44} to both sides of the equation.

$$26x=\frac{4459}{44}$$

Divide both sides by 26.

$$x=\frac{343}{88}$$

The system is now solved.

$$x=\frac{343}{88},y=-\frac{883}{44}$$