Linear equations and applications

Example 1: Solve ${3(x-2)+5=2x+7}$.

Solution:

Remove the parentheses, and combine like terms on the left side:

$\begin{align*}3x-6+5&=2x+7\\3x-1&=2x+7\end{align*}$

Subtract ${2x}$ from both sides. Then add ${1}$ to both sides:

$\begin{align*}x-1&=7\\x&=8\end{align*}$

Check: The left side is ${3(8-2)+5=3(6)+5=23}$. The right side is ${2(8)+7=23}$. They are equal, so the solution is ${8}$.

Example 2: Solve ${\dfrac{x}{3}-\dfrac{x-1}{4}=2}$.

Solution:

The LCD of ${3}$ and ${4}$ is ${12}$. Multiply every term on both sides by ${12}$:

$\begin{align*}12\cdot\dfrac{x}{3}-12\cdot\dfrac{x-1}{4}&=12\cdot 2\\4x-3(x-1)&=24\end{align*}$

Keep the parentheses around ${x-1}$, so the minus sign applies to both of its terms:

$\begin{align*}4x-3x+3&=24\\x+3&=24\\x&=21\end{align*}$

Check: ${\dfrac{21}{3}-\dfrac{20}{4}=7-5=2}$. It checks.

Example 3: Solve each equation.

(a) ${2(x+3)=2x+5}$

(b) ${3x+6=3(x+2)}$

Solution:

(a) Remove the parentheses, then subtract ${2x}$ from both sides:

$\begin{align*}2x+6&=2x+5\\6&=5\end{align*}$

This is false, whatever ${x}$ is. So the equation has no solution.

(b) Remove the parentheses, then subtract ${3x}$ from both sides:

$\begin{align*}3x+6&=3x+6\\6&=6\end{align*}$

This is always true. So every real number is a solution.

Example 4: Solve each formula for the given variable.

(a) ${P=2l+2w}$ for ${w}$. (This is the perimeter ${P}$ of a rectangle with length ${l}$ and width ${w}$.)

(b) ${A=P(1+rt)}$ for ${r}$. (This is the amount ${A}$ of an investment of ${P}$ dollars at simple interest rate ${r}$ for ${t}$ years.)

Solution:

(a) Subtract ${2l}$ from both sides. Then divide both sides by ${2}$:

$\begin{align*}P-2l&=2w\\w&=\dfrac{P-2l}{2}\end{align*}$

(b) Remove the parentheses. Then get the term with ${r}$ alone:

$\begin{align*}A&=P+Prt\\A-P&=Prt\\r&=\dfrac{A-P}{Pt}\end{align*}$

Example 5: A student invests ${10{,}000}$ dollars. Part of it earns ${4\%}$ interest per year, and the rest earns ${6\%}$. After one year, the total interest is ${520}$ dollars. How much was invested at each rate?

Solution:

Let ${x}$ be the amount, in dollars, invested at ${4\%}$. The rest, ${10{,}000-x}$ dollars, is invested at ${6\%}$.

The interest on each part is the rate times the amount. As decimals, ${4\%=0.04}$ and ${6\%=0.06}$. The two interests add up to ${520}$:

${0.04x+0.06(10{,}000-x)=520}$

Remove the parentheses, and combine like terms:

$\begin{align*}0.04x+600-0.06x&=520\\-0.02x+600&=520\\-0.02x&=-80\\x&=4000\end{align*}$

So ${4000}$ dollars is invested at ${4\%}$, and ${10{,}000-4000=6000}$ dollars at ${6\%}$.

Check: Add the interest from the two parts:

$\begin{align*}&0.04(4000)+0.06(6000)\\&=160+360\\&=520\end{align*}$

It checks.

Example 6: A freight train leaves a station at ${60}$ miles per hour (mi/h). Two hours later, a car leaves the same station, driving along the same route at ${80}$ mi/h. How long does the car take to catch up with the train?

Solution:

Both start at the station and move in the same direction. The car starts ${2}$ hours late, but it is faster, so it slowly closes the gap. It catches up when both have gone the same distance from the station.

For motion at a steady speed, distance is rate times time: ${d=rt}$.

Let ${t}$ be the car's driving time, in hours, until it catches up. The train started ${2}$ hours earlier, so it has been moving for ${t+2}$ hours. Put the facts in a table:

	Rate (mi/h)	Time (h)	Distance (mi)
Train	${60}$	${t+2}$	${60(t+2)}$
Car	${80}$	${t}$	${80t}$

The distances are equal when the car catches up:

$\begin{align*}80t&=60(t+2)\\80t&=60t+120\\20t&=120\\t&=6\end{align*}$

The car catches up after ${6}$ hours of driving.

Check: The car goes ${80(6)=480}$ miles. The train goes ${60(8)=480}$ miles. The distances match.

Example 7: The length of a rectangle is ${3}$ meters more than twice its width. Its perimeter is ${54}$ meters. Find its length and width.

Solution:

Let ${w}$ be the width, in meters. Then the length is ${2w+3}$.

The perimeter is twice the length plus twice the width:

$\begin{align*}2(2w+3)+2w&=54\\4w+6+2w&=54\\6w&=48\\w&=8\end{align*}$

The width is ${8}$ m, and the length is ${2(8)+3=19}$ m.

Check: ${2(19)+2(8)=38+16=54}$. It checks.