Solve {l}{x=20t}{x=30(t-5)}

$$\left. \begin{array} { l } { x = 20 t } \\ { x = 30 ( t - 5 ) } \end{array} \right.$$

Solve for x, t

$$x=300$$

$$t=15$$

Steps Using Substitution

Consider the first equation. Subtract 20t from both sides.

$$x-20t=0$$

Consider the second equation. Use the distributive property to multiply 30 by t-5.

$$x=30t-150$$

Subtract 30t from both sides.

$$x-30t=-150$$

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.

$$x-20t=0,x-30t=-150$$

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

$$x-20t=0$$

Add 20t to both sides of the equation.

$$x=20t$$

Substitute 20t for x in the other equation, x-30t=-150.

$$20t-30t=-150$$

Add 20t to -30t.

$$-10t=-150$$

Divide both sides by -10.

$$t=15$$

Substitute 15 for t in x=20t. Because the resulting equation contains only one variable, you can solve for x directly.

$$x=20\times 15$$

Multiply 20 times 15.

$$x=300$$

The system is now solved.

$$x=300,t=15$$

Steps Using Matrices

Consider the first equation. Subtract 20t from both sides.

$$x-20t=0$$

Consider the second equation. Use the distributive property to multiply 30 by t-5.

$$x=30t-150$$

Subtract 30t from both sides.

$$x-30t=-150$$

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

$$x-20t=0,x-30t=-150$$

Write the equations in matrix form.

$$\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right)\left(\begin{matrix}x\\t\end{matrix}\right)=\left(\begin{matrix}0\\-150\end{matrix}\right)$$

Left multiply the equation by the inverse matrix of \left(\begin{matrix}1&-20\\1&-30\end{matrix}\right).

$$inverse(\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right))\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right)\left(\begin{matrix}x\\t\end{matrix}\right)=inverse(\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right))\left(\begin{matrix}0\\-150\end{matrix}\right)$$

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

$$\left(\begin{matrix}1&0\\0&1\end{matrix}\right)\left(\begin{matrix}x\\t\end{matrix}\right)=inverse(\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right))\left(\begin{matrix}0\\-150\end{matrix}\right)$$

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

$$\left(\begin{matrix}x\\t\end{matrix}\right)=inverse(\left(\begin{matrix}1&-20\\1&-30\end{matrix}\right))\left(\begin{matrix}0\\-150\end{matrix}\right)$$

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.

$$\left(\begin{matrix}x\\t\end{matrix}\right)=\left(\begin{matrix}-\frac{30}{-30-\left(-20\right)}&-\frac{-20}{-30-\left(-20\right)}\\-\frac{1}{-30-\left(-20\right)}&\frac{1}{-30-\left(-20\right)}\end{matrix}\right)\left(\begin{matrix}0\\-150\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\t\end{matrix}\right)=\left(\begin{matrix}3&-2\\\frac{1}{10}&-\frac{1}{10}\end{matrix}\right)\left(\begin{matrix}0\\-150\end{matrix}\right)$$

Multiply the matrices.

$$\left(\begin{matrix}x\\t\end{matrix}\right)=\left(\begin{matrix}-2\left(-150\right)\\-\frac{1}{10}\left(-150\right)\end{matrix}\right)$$

Do the arithmetic.

$$\left(\begin{matrix}x\\t\end{matrix}\right)=\left(\begin{matrix}300\\15\end{matrix}\right)$$

Extract the matrix elements x and t.

$$x=300,t=15$$

Steps Using Elimination

Consider the first equation. Subtract 20t from both sides.

$$x-20t=0$$

Consider the second equation. Use the distributive property to multiply 30 by t-5.

$$x=30t-150$$

Subtract 30t from both sides.

$$x-30t=-150$$

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.

$$x-20t=0,x-30t=-150$$

Subtract x-30t=-150 from x-20t=0 by subtracting like terms on each side of the equal sign.

$$x-x-20t+30t=150$$

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

$$-20t+30t=150$$

Add -20t to 30t.

$$10t=150$$

Divide both sides by 10.

$$t=15$$

Substitute 15 for t in x-30t=-150. Because the resulting equation contains only one variable, you can solve for x directly.

$$x-30\times 15=-150$$

Multiply -30 times 15.

$$x-450=-150$$

Add 450 to both sides of the equation.

$$x=300$$

The system is now solved.

$$x=300,t=15$$