Solve {l}{6x+3y=17}{17+64x=y}

$$\left\{ \begin{array} { l } { 6 x + 3 y = 17 } \\ { 17 + 64 x = y } \end{array} \right.$$

Solve for x, y

$$x=-\frac{17}{99}\approx -0.171717172$$

$$y = \frac{595}{99} = 6\frac{1}{99} \approx 6.01010101$$

Steps Using Substitution

Consider the second equation. Subtract y from both sides.

$$17+64x-y=0$$

Subtract 17 from both sides. Anything subtracted from zero gives its negation.

$$64x-y=-17$$

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.

$$6x+3y=17,64x-y=-17$$

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

$$6x+3y=17$$

Subtract 3y from both sides of the equation.

$$6x=-3y+17$$

Divide both sides by 6.

$$x=\frac{1}{6}\left(-3y+17\right)$$

Multiply \frac{1}{6} times -3y+17.

$$x=-\frac{1}{2}y+\frac{17}{6}$$

Substitute -\frac{y}{2}+\frac{17}{6} for x in the other equation, 64x-y=-17.

$$64\left(-\frac{1}{2}y+\frac{17}{6}\right)-y=-17$$

Multiply 64 times -\frac{y}{2}+\frac{17}{6}.

$$-32y+\frac{544}{3}-y=-17$$

Add -32y to -y.

$$-33y+\frac{544}{3}=-17$$

Subtract \frac{544}{3} from both sides of the equation.

$$-33y=-\frac{595}{3}$$

Divide both sides by -33.

$$y=\frac{595}{99}$$

Substitute \frac{595}{99} for y in x=-\frac{1}{2}y+\frac{17}{6}. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=-\frac{1}{2}\times \frac{595}{99}+\frac{17}{6}$$

Multiply -\frac{1}{2} times \frac{595}{99} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{595}{198}+\frac{17}{6}$$

Add \frac{17}{6} to -\frac{595}{198} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{17}{99}$$

The system is now solved.

$$x=-\frac{17}{99},y=\frac{595}{99}$$

Steps Using Matrices

Consider the second equation. Subtract y from both sides.

$$17+64x-y=0$$

Subtract 17 from both sides. Anything subtracted from zero gives its negation.

$$64x-y=-17$$

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

$$6x+3y=17,64x-y=-17$$

Write the equations in matrix form.

$$\left(\begin{matrix}6&3\\64&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}17\\-17\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}6&3\\64&-1\end{matrix}\right).

$$inverse(\left(\begin{matrix}6&3\\64&-1\end{matrix}\right))\left(\begin{matrix}6&3\\64&-1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\64&-1\end{matrix}\right))\left(\begin{matrix}17\\-17\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\64&-1\end{matrix}\right))\left(\begin{matrix}17\\-17\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\y\end{matrix}\right)=inverse(\left(\begin{matrix}6&3\\64&-1\end{matrix}\right))\left(\begin{matrix}17\\-17\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{1}{6\left(-1\right)-3\times 64}&-\frac{3}{6\left(-1\right)-3\times 64}\\-\frac{64}{6\left(-1\right)-3\times 64}&\frac{6}{6\left(-1\right)-3\times 64}\end{matrix}\right)\left(\begin{matrix}17\\-17\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{198}&\frac{1}{66}\\\frac{32}{99}&-\frac{1}{33}\end{matrix}\right)\left(\begin{matrix}17\\-17\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}\frac{1}{198}\times 17+\frac{1}{66}\left(-17\right)\\\frac{32}{99}\times 17-\frac{1}{33}\left(-17\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\y\end{matrix}\right)=\left(\begin{matrix}-\frac{17}{99}\\\frac{595}{99}\end{matrix}\right)$$

Extract the matrix elements x and y.

$$x=-\frac{17}{99},y=\frac{595}{99}$$

Steps Using Elimination

Consider the second equation. Subtract y from both sides.

$$17+64x-y=0$$

Subtract 17 from both sides. Anything subtracted from zero gives its negation.

$$64x-y=-17$$

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.

$$6x+3y=17,64x-y=-17$$

To make 6x and 64x equal, multiply all terms on each side of the first equation by 64 and all terms on each side of the second by 6.

$$64\times 6x+64\times 3y=64\times 17,6\times 64x+6\left(-1\right)y=6\left(-17\right)$$

Simplify.

$$384x+192y=1088,384x-6y=-102$$

Subtract 384x-6y=-102 from 384x+192y=1088 by subtracting like terms on each side of the equal sign.

$$384x-384x+192y+6y=1088+102$$

Add 384x to -384x. Terms 384x and -384x cancel out, leaving an equation with only one variable that can be solved.

$$192y+6y=1088+102$$

Add 192y to 6y.

$$198y=1088+102$$

Add 1088 to 102.

$$198y=1190$$

Divide both sides by 198.

$$y=\frac{595}{99}$$

Substitute \frac{595}{99} for y in 64x-y=-17. Because the resulting equation contains only one variable, you can solve for x directly.

$$64x-\frac{595}{99}=-17$$

Add \frac{595}{99} to both sides of the equation.

$$64x=-\frac{1088}{99}$$

Divide both sides by 64.

$$x=-\frac{17}{99}$$

The system is now solved.

$$x=-\frac{17}{99},y=\frac{595}{99}$$