Solve {l}{5x=7z}{8x=9z}

$$\left. \begin{array} { l } { 5 x = 7 z } \\ { 8 x = 9 z } \end{array} \right.$$

Solve for x, z

$$x=0$$

$$z=0$$

Steps Using Substitution

Consider the first equation. Subtract 7z from both sides.

$$5x-7z=0$$

Consider the second equation. Subtract 9z from both sides.

$$8x-9z=0$$

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.

$$5x-7z=0,8x-9z=0$$

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

$$5x-7z=0$$

Add 7z to both sides of the equation.

$$5x=7z$$

Divide both sides by 5.

$$x=\frac{1}{5}\times 7z$$

Multiply \frac{1}{5} times 7z.

$$x=\frac{7}{5}z$$

Substitute \frac{7z}{5} for x in the other equation, 8x-9z=0.

$$8\times \frac{7}{5}z-9z=0$$

Multiply 8 times \frac{7z}{5}.

$$\frac{56}{5}z-9z=0$$

Add \frac{56z}{5} to -9z.

$$\frac{11}{5}z=0$$

Divide both sides of the equation by \frac{11}{5}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$z=0$$

Substitute 0 for z in x=\frac{7}{5}z. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=0$$

The system is now solved.

$$x=0,z=0$$

Steps Using Matrices

Consider the first equation. Subtract 7z from both sides.

$$5x-7z=0$$

Consider the second equation. Subtract 9z from both sides.

$$8x-9z=0$$

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

$$5x-7z=0,8x-9z=0$$

Write the equations in matrix form.

$$\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}5&-7\\8&-9\end{matrix}\right).

$$inverse(\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right))\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\z\end{matrix}\right)=inverse(\left(\begin{matrix}5&-7\\8&-9\end{matrix}\right))\left(\begin{matrix}0\\0\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{5\left(-9\right)-\left(-7\times 8\right)}&-\frac{-7}{5\left(-9\right)-\left(-7\times 8\right)}\\-\frac{8}{5\left(-9\right)-\left(-7\times 8\right)}&\frac{5}{5\left(-9\right)-\left(-7\times 8\right)}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}-\frac{9}{11}&\frac{7}{11}\\-\frac{8}{11}&\frac{5}{11}\end{matrix}\right)\left(\begin{matrix}0\\0\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\z\end{matrix}\right)=\left(\begin{matrix}0\\0\end{matrix}\right)$$

Extract the matrix elements x and z.

$$x=0,z=0$$

Steps Using Elimination

Consider the first equation. Subtract 7z from both sides.

$$5x-7z=0$$

Consider the second equation. Subtract 9z from both sides.

$$8x-9z=0$$

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.

$$5x-7z=0,8x-9z=0$$

To make 5x and 8x equal, multiply all terms on each side of the first equation by 8 and all terms on each side of the second by 5.

$$8\times 5x+8\left(-7\right)z=0,5\times 8x+5\left(-9\right)z=0$$

Simplify.

$$40x-56z=0,40x-45z=0$$

Subtract 40x-45z=0 from 40x-56z=0 by subtracting like terms on each side of the equal sign.

$$40x-40x-56z+45z=0$$

Add 40x to -40x. Terms 40x and -40x cancel out, leaving an equation with only one variable that can be solved.

$$-56z+45z=0$$

Add -56z to 45z.

$$-11z=0$$

Divide both sides by -11.

$$z=0$$

Substitute 0 for z in 8x-9z=0. Because the resulting equation contains only one variable, you can solve for x directly.

$$8x=0$$

Divide both sides by 8.

$$x=0$$

The system is now solved.

$$x=0,z=0$$