Two-Step Equations Word Problems Practice

1. Cookies are on sale! Today each cookie costs ‍$0.75 less than the normal price. Right now if you buy ‍7 of them it will only cost you $2.80!

   Write an equation to determine the normal price of each cookie \((c)\).

   Find the normal price of each cookie.

2. John read the first ‍114 pages of a novel, which was ‍3 pages less than ⅓ of the novel.

   Write an equation to determine the total number of pages \((p)\) in the novel.

   Find the total number of pages in the novel.

3. In winter, the price of apples suddenly went up by ‍$0.75 per pound. Sam bought ‍3 pounds of apples at the new price for a total of ‍$5.88.

   Write an equation to determine the original price per pound \((p)\).

   Find the original price per pound.

4. The perimeter of a rectangle is ‍34 units. Its width is ‍6.5 units.

   Write an equation to determine the length ‍\((l)\) of the rectangle.

   Find the length of the rectangle.

Source: Khan Academy, https://www.khanacademy.org/math/algebra-home/alg-basic-eq-ineq/alg-linear-eq-word-probs/e/linear-equation-world-problems-2
This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 License.

1. Let ‍\(c\) be the normal price of each cookie.

   The current price for each cookie is ‍\(c-0.75\).

   The current cost of ‍7 cookies is ‍\(7(c-0.75)\).
   

   Since the current cost of ‍7 cookies is ‍$2.80, let's set this equal to ‍2.8:
   

   \(7(c-0.75)=2.8\)

   Now, let's solve the equation to find the normal price of each cookie (c).

   \(\begin{aligned}7(c-0.75)&=2.8\\&\\
   \dfrac{7(c-0.75)}{{7}}&=\dfrac{2.8}{{7}}&&\text{divide both sides by ${7}$}\\
   \\
   c-0.75&=0.4\\
   \\
   c-{0.75}{+0.75}&=0.4{+0.75}&&{\text{add }} {0.75} \text{ to both sides}\\
   \\
   c&=1.15\end{aligned}\)

   The equation is ‍\(7(c-0.75)=2.8\).

   The normal price of each cookie is ‍\(\$1.15\).

2. Let ‍\(p\) be the total number of pages in the novel.

   We can represent \(\dfrac{1}{3}\) of the novel as \(\dfrac{1}{3}p\). John has read ‍3 fewer pages than \(\dfrac{1}{3}p\).

   He has read \(\dfrac{1}{3}p-3\)pages.

   Since he has read 114 pages, let's set this equal to 114:

   Now, let's solve the equation to find the total number of pages \((p)\) in the novel.

   \(\begin{aligned}
   \dfrac{1}3p-3&=114\\
   \\
   \dfrac{1}3p-3{+3}&=114{+3}&&{\text{add }3} \text{ to each side}\\
   \\
   \dfrac{1}3p&=117\\
   \\
   \dfrac{\dfrac{1}3p}{{\dfrac{1}3}}&=\dfrac{117}{{\dfrac{1}3}}&&\text{divide each side by ${\dfrac{1}3}$}\\
   \\
   p&=351\end{aligned}\)

   The equation is \(\dfrac{1}{3}p-3 = 114\).
   

   The novel has a total of ‍351 pages.

3. Let \(p\)  be the original price per pound of apples.

   The new price is \(p+{0.75}\) dollars per pound. Sam bought 3 pounds of apples.

   Sam's total cost was \(3(p+0.75)\).

   Since his total cost was \(\$5.88\), let's set this equal to 5.88:

   Now, let's solve the equation to find the original price per pound \((p)\).

   \(\begin{aligned}3(p+0.75)&=5.88\\&\\
   \dfrac{3(p+{0.75})}{3}&=\dfrac{5.88}{3}&&\text{divide both sides by $3$}\\
   \\
   p+{0.75}&=1.96\\
   \\
   p+{0.75}{-0.75}&=1.96{-0.75}&&{\text{subtract }} {0.75} \text{ from both sides}\\
   \\
   p&=1.21\end{aligned}\)

   The equation is \(3(p+0.75)=5.88\).

   The original price of the apples was $1.21 per pound.
   

4. Let \(l\) be the length of the rectangle.

   The perimeter is equal to \(2l+2w\). Let's substitute in the width of 6.5:

   \(\qquad\begin{aligned}&2l+2w\\
   =&2l+2(6.5)\\
   =&2l+13\end{aligned}\)
   

   The perimeter of the rectangle is \(2l+{13}\).

   Since the perimeter equals 34 units, let's set this equal to 34:

   Now, let's solve the equation to find the length of the rectangle \((l)\).

   \(\begin{aligned}
   2l+13&=34\\
   \\
   2l+13{-13}&=34{-13}&&{\text{subtract }13} \text{ from each side}\\
   \\
   2l&=21\\
   \\
   \dfrac{2l}{{2}}&=\dfrac{21}{{2}}&&\text{divide each side by ${2}$}\\
   \\
   l&=10.5\end{aligned}\)
   

   The equation is \(2l+13=34\).

   The length of the rectangle is 10.5 units.