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In the last example, we could have made the left side the “variable” side, but it would have led to a negative coefficient on the variable term. (Try it!) While we could work with the negative, there is less chance of errors when working with positives. The strategy outlined above helps avoid the negatives!
To solve an equation with fractions, we just follow the steps of our strategy to get the solution!
Solve: $\frac{5}{4}x+6=\frac{1}{4}x-2.$
Since $\frac{5}{4}>\frac{1}{4}$ , make the left side the “variable” side and the right side the “constant” side.
Subtract $\frac{1}{4}x$ from both sides. | |
Combine like terms. | |
Subtract $6$ from both sides. | |
Simplify. | |
$\begin{array}{cccccc}\text{Check:}\hfill & & & \hfill \frac{5}{4}x+6& =\hfill & \frac{1}{4}x-2\hfill \\ \text{Let}\phantom{\rule{0.2em}{0ex}}x=\mathrm{-8}.\hfill & & & \hfill \frac{5}{4}(\mathrm{-8})+6& \stackrel{?}{=}\hfill & \frac{1}{4}(\mathrm{-8})-2\hfill \\ & & & \hfill \mathrm{-10}+6& \stackrel{?}{=}\hfill & \mathrm{-2}-2\hfill \\ & & & \hfill \mathrm{-4}& =\hfill & \mathrm{-4}\u2713\hfill \end{array}$ |
We will use the same strategy to find the solution for an equation with decimals.
Solve: $7.8x+4=5.4x-8.$
Since $7.8>5.4$ , make the left side the “variable” side and the right side the “constant” side.
Subtract $5.4x$ from both sides. | ||
Combine like terms. | ||
Subtract $4$ from both sides. | ||
Simplify. | ||
Use the Division Propery of Equality. | ||
Simplify. | ||
Check: | ||
Let $x=\mathrm{-5}$ . | ||
Solve Equations with Constants on Both Sides
In the following exercises, solve the following equations with constants on both sides.
$9x-3=60$
$14w+5=117$
$2a+8=\mathrm{-28}$
$\mathrm{-62}=8n-6$
$35=\mathrm{-13}y+9$
$\mathrm{-12}p-9=9$
Solve Equations with Variables on Both Sides
In the following exercises, solve the following equations with variables on both sides.
$19z=18z-7$
$9x+36=15x$
$c=\mathrm{-3}c-20$
$9q=44-2q$
$6y+\frac{1}{2}=5y$
$\mathrm{-18}a-8=\mathrm{-22}a$
Solve Equations with Variables and Constants on Both Sides
In the following exercises, solve the following equations with variables and constants on both sides.
$8x-15=7x+3$
$26+13d=14d+11$
$2p-1=4p-33$
$4a+5=\text{\u2212}a-40$
$5y-30=\mathrm{-5}y+30$
$7s+12=5+4s$
$2z-6=23-z$
$\frac{5}{3}c-3=\frac{2}{3}c-16$
$8-\frac{2}{5}q=\frac{3}{5}q+6$
$\frac{4}{3}n+9=\frac{1}{3}n-9$
$\frac{1}{4}y+7=\frac{3}{4}y-3$
$14n+8.25=9n+19.60$
$2.4w-100=0.8w+28$
$5.6r+13.1=3.5r+57.2$
Concert tickets At a school concert the total value of tickets sold was $1506. Student tickets sold for $6 and adult tickets sold for $9. The number of adult tickets sold was 5 less than 3 times the number of student tickets. Find the number of student tickets sold, s , by solving the equation $6s+27s-45=1506$ .
Making a fence Jovani has 150 feet of fencing to make a rectangular garden in his backyard. He wants the length to be 15 feet more than the width. Find the width, w , by solving the equation $150=2w+30+2w$ .
30 feet
Solve the equation $\frac{6}{5}y-8=\frac{1}{5}y+7$ explaining all the steps of your solution as in the examples in this section.
Solve the equation $10x+14=\mathrm{-2}x+38$ explaining all the steps of your solution as in the examples in this section.
$x=2$ Justifications will vary.
When solving an equation with variables on both sides, why is it usually better to choose the side with the larger coefficient of $x$ to be the “variable” side?
Is $x=\mathrm{-2}$ a solution to the equation $5-2x=\mathrm{-4}x+1$ ? How do you know?
Yes. Justifications will vary.
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?
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