Solve {l}{52x+38y=0}{47x+61y=99}

$$\left\{ \begin{array} { l } { 52 x + 38 y = 0 } \\ { 47 x + 61 y = 99 } \end{array} \right.$$

Solve for x, y

$$x = -\frac{19}{7} = -2\frac{5}{7} \approx -2.714285714$$

$$y = \frac{26}{7} = 3\frac{5}{7} \approx 3.714285714$$

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.

$$52x+38y=0,47x+61y=99$$

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

$$52x+38y=0$$

Subtract 38y from both sides of the equation.

$$52x=-38y$$

Divide both sides by 52.

$$x=\frac{1}{52}\left(-38\right)y$$

Multiply \frac{1}{52} times -38y.

$$x=-\frac{19}{26}y$$

Substitute -\frac{19y}{26} for x in the other equation, 47x+61y=99.

$$47\left(-\frac{19}{26}\right)y+61y=99$$

Multiply 47 times -\frac{19y}{26}.

$$-\frac{893}{26}y+61y=99$$

Add -\frac{893y}{26} to 61y.

$$\frac{693}{26}y=99$$

Divide both sides of the equation by \frac{693}{26}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$y=\frac{26}{7}$$

Substitute \frac{26}{7} for y in x=-\frac{19}{26}y. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{19}{26}\times \frac{26}{7}$$

Multiply -\frac{19}{26} times \frac{26}{7} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{19}{7}$$

The system is now solved.

$$x=-\frac{19}{7},y=\frac{26}{7}$$

Steps Using Matrices

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

$$52x+38y=0,47x+61y=99$$

Write the equations in matrix form.

$$\left(\begin{matrix}52&38\\47&61\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}0\\99\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}52&38\\47&61\end{matrix}\right).

$$inverse(\left(\begin{matrix}52&38\\47&61\end{matrix}\right))\left(\begin{matrix}52&38\\47&61\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}52&38\\47&61\end{matrix}\right))\left(\begin{matrix}0\\99\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}52&38\\47&61\end{matrix}\right))\left(\begin{matrix}0\\99\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}52&38\\47&61\end{matrix}\right))\left(\begin{matrix}0\\99\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{61}{52\times 61-38\times 47}&-\frac{38}{52\times 61-38\times 47}\\-\frac{47}{52\times 61-38\times 47}&\frac{52}{52\times 61-38\times 47}\end{matrix}\right)\left(\begin{matrix}0\\99\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{61}{1386}&-\frac{19}{693}\\-\frac{47}{1386}&\frac{26}{693}\end{matrix}\right)\left(\begin{matrix}0\\99\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{693}\times 99\\\frac{26}{693}\times 99\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{19}{7}\\\frac{26}{7}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{19}{7},y=\frac{26}{7}$$

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.

$$52x+38y=0,47x+61y=99$$

To make 52x and 47x equal, multiply all terms on each side of the first equation by 47 and all terms on each side of the second by 52.

$$47\times 52x+47\times 38y=0,52\times 47x+52\times 61y=52\times 99$$

Simplify.

$$2444x+1786y=0,2444x+3172y=5148$$

Subtract 2444x+3172y=5148 from 2444x+1786y=0 by subtracting like terms on each side of the equal sign.

$$2444x-2444x+1786y-3172y=-5148$$

Add 2444x to -2444x. Terms 2444x and -2444x cancel out, leaving an equation with only one variable that can be solved.

$$1786y-3172y=-5148$$

Add 1786y to -3172y.

$$-1386y=-5148$$

Divide both sides by -1386.

$$y=\frac{26}{7}$$

Substitute \frac{26}{7} for y in 47x+61y=99. Because the resulting equation contains only one variable, you can solve for x directly.

$$47x+61\times \frac{26}{7}=99$$

Multiply 61 times \frac{26}{7}.

$$47x+\frac{1586}{7}=99$$

Subtract \frac{1586}{7} from both sides of the equation.

$$47x=-\frac{893}{7}$$

Divide both sides by 47.

$$x=-\frac{19}{7}$$

The system is now solved.

$$x=-\frac{19}{7},y=\frac{26}{7}$$