Tuition increases on November 15th - Enroll soon to lock in current pricing!

Chapter 5: Equations and Inequalities
5.3 Solving Linear Equations II
C
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
5.3 Solving Linear Equations II
Problems
Problem 5.12
4 Jump to Solution

In this problem, we solve the equation $\color[rgb]{0.11,0.21,0.37}8t + 9 = 65$.

(a) Isolate the $\color[rgb]{0.11,0.21,0.37}8t$ by subtracting an appropriate constant from both sides.

(b) Solve the resulting equation for $\color[rgb]{0.11,0.21,0.37}t$.

Problem 5.13
4 Jump to Solution

In this problem, we solve the equation $\color[rgb]{0.11,0.21,0.37}7j - 4 + 3j = 6 + 2j - 4j - 8$.

(a) Simplify both sides of the equation by combining like terms.

(b) Add an expression to both sides of your equation from part (a) to give an equation in which no variables are on the right-hand side.

(c) Solve the equation resulting from part (b).

(d) Check your answer! Substitute your value of $\color[rgb]{0.11,0.21,0.37}j$ into the original equation. If it doesn’t work, then do the problem again.

Problem 5.14
4 Jump to Solution

Solve the following equations:

(a) $\color[rgb]{0.11,0.21,0.37}8k - 13\frac25 = -12\frac{1}{25}\phantom{\dfrac12}$

(b) $\color[rgb]{0.11,0.21,0.37}4(t-7) = 3(2t+3)\phantom{\dfrac12}$

(c) $\color[rgb]{0.11,0.21,0.37}\dfrac{2r-7}{9} = 3$

(d) $\color[rgb]{0.11,0.21,0.37} \dfrac{3x+4}{5}=\dfrac{2x-8}{7}$

Problem 5.15
4 Jump to Solution

Solve the following equations:

(a) $\color[rgb]{0.11,0.21,0.37} \dfrac95 -\dfrac{2x}{3} = \dfrac{6x}{5} + \dfrac73$

(b) $\color[rgb]{0.11,0.21,0.37}\dfrac{4-7t}{6} = \dfrac{t}{8} + 2$

Problem 5.16
4 Jump to Solution

(a) Find all values of $\color[rgb]{0.11,0.21,0.37}w$ that satisfy $\color[rgb]{0.11,0.21,0.37}5w + 3- 2w = w - 8 + 2w - 3$.

(b) Find all values of $\color[rgb]{0.11,0.21,0.37}z$ that satisfy $\color[rgb]{0.11,0.21,0.37}2z - 8 - 5z = 2 - 3z -10$.

Problem 5.17
4 Jump to Solution

For what value of $\color[rgb]{0.11,0.21,0.37}c$ do the equations $\color[rgb]{0.11,0.21,0.37}2y - 5 = 17$ and $\color[rgb]{0.11,0.21,0.37}cy - 8 = 36$ have the same solution for $\color[rgb]{0.11,0.21,0.37}y$?

In the last section, we used addition and subtraction to solve some equations, and used multiplication and division to solve others. To solve most linear equations, however, we’ll have to use a combination of these tactics.

Problem 5.12
t
V

Solve the equation $\color[rgb]{1,1,1}8t + 9 = 65$.

Solution for Problem 5.12: This equation doesn’t look exactly like any of the equations we already know how to solve. It may not be obvious immediately how to isolate $t$. However, we can isolate $8t$ by subtracting 9 from both sides:

$\begin{array}{@{\extracolsep{-0.09in}}rcrcr} 8t&+&9 &=& 65\\\rule{0pt}{0.15in} &-&9 &=& -9\vspace*{0.02in}\\\hline\rule{0pt}{0.17in} 8t&&&=&56 \end{array}$

Now we have an equation we know how to solve! We divide both sides by 8 to find $t=7$.

We can check our work by substituting this value for $t$ back into our original equation. We find that $8(7) + 9 = 65$, so our answer works.

We didn’t have to add first when we solved this equation. We could have divided first: \[\frac{8t+9}{8} = \frac{65}{8}.\]We can then distribute on the left side. Since \[\frac{8t+9}{8} = \frac{8t}{8} + \frac{9}{8} = t + \frac{9}{8},\]we have \[t + \frac98 = \frac{65}{8}.\]We then subtract $\dfrac{9}{8}$ from both sides of this equation to get $t = \dfrac{65}{8} - \dfrac{9}{8} = \dfrac{56}{8} = 7$, as before. □

The equation in Problem 5.12 is not exactly like any of the equations we solved in the previous section. However, we were still able to solve it with the same tools.

l
Concept:

When solving an equation that isn’t exactly like an equation you have solved before, try to manipulate it into a form you already know how to deal with.

See if you can apply this strategy to the following problem.

Problem 5.13
t
V

Solve the equation $\color[rgb]{1,1,1}7j - 4 + 3j = 6 + 2j - 4j - 8$.

Solution for Problem 5.13: Our first step is to simplify both sides of the equation. By grouping like terms, the left-hand side of the original equation becomes \[7j-4+3j = (7j+3j) - 4 = 10j - 4.\]The right-hand side of the original equation becomes \[6+2j-4j-8= (2j-4j) + (6-8) = -2j-2.\]

Combining these results simplifies the original equation to \[10j-4 = -2j-2.\]We haven’t solved any equations in which the variable appears on both sides. We know how to handle an equation if the variable only appears on one side. So, we add $2j$ to both sides to eliminate the variable from the right-hand side:

$\begin{array}{@{\extracolsep{-0.09in}}rcrcrcr} 10j&-&4 &=& -2j&-&2\\\rule{0pt}{0.15in} +2j&& &=& +2j&&\vspace*{0.02in}\\\hline\rule{0pt}{0.17in} 12j&-&4&=&&-&2 \end{array}$

Now we have an equation we know how to solve! We add 4 to both sides to get $12j = 2$. We then divide by 12 to find $j = \frac{2}{12} = \frac16$.

We now have another strategy for solving linear equations.

l
Concept:

If the variable appears on both sides of the equation, we can use addition and subtraction to get all terms with the variable on the same side of the equation.

Similarly, we use addition and subtraction to get all the constant terms on the other side of the equation.

Here’s a little more practice.

Problem 5.14
t
V

Solve the following equations:

(a) $\color[rgb]{1,1,1}8k - 13\frac25 = -12\frac{1}{25}\phantom{\dfrac12}$

(b) $\color[rgb]{1,1,1}4(t-7) = 3(2t+3)\phantom{\dfrac12}$

(c) $\color[rgb]{1,1,1}\dfrac{2r-7}{9} = 3$

(d) $\color[rgb]{1,1,1} \dfrac{3x+4}{5}=\dfrac{2x-8}{7}$

Solution for Problem 5.14:

(a) Adding 13$\frac25$ to both sides leaves the variable term on the left while putting all the constant terms on the right: \[8k = -12\frac{1}{25} + 13\frac25.\]Simplifying the right-hand side gives \[ -12\frac{1}{25} + 13\frac25 = (-12+13) +\left(-\frac{1}{25} + \frac25\right) = 1\frac{9}{25}, \]so we now have \[8k = 1\frac{9}{25}.\]Multiplying both sides by $\frac18$ (which is the same as dividing both sides by 8) gives \[k = \frac18\cdot 1\frac{9}{25} = \frac18\cdot\frac{34}{25} = \frac{34}{200} =\frac{17}{100}.\]

(b) First, we use the distributive property to expand both sides: \[4\cdot t - 4\cdot 7 = 3\cdot2t + 3\cdot 3.\]Simplifying both sides gives \[4t - 28 = 6t + 9.\]Next, we get all the terms with $t$ on one side of the equation and all the constants on the other side. Subtracting $4t$ from both sides gives $-28 = 2t +9$. Subtracting 9 from both sides gives $-37 = 2t$. Finally, dividing both sides by 2 gives $t=-\frac{37}{2}$.

(c) First, make sure you see why adding 7 to $\frac{2r-7}{9}$ doesn’t “cancel the $-7$.” This is because $\frac{2r-7}{9} + 7$ equals $\frac{2r}{9} - \frac79 + 7$, which is $\frac{2r}{9} +\frac{56}{9}$. There’s still a constant term; the $\frac{2r}{9}$ term is not yet isolated.

Since $\frac{2r-7}{9}$ equals $\frac{2r}{9} -\frac79$, we add $\frac79$ to both sides of \[\frac{2r}{9} -\frac79 = 3\]to eliminate the constant on the left side and isolate $\frac{2r}{9}$. Doing so gives us \[\frac{2r}{9} = 3+\frac79= \frac{34}{9}.\]Multiplying both sides of $\frac{2r}{9}=\frac{34}{9}$ by $\frac92$ gives $r = \frac{34}{9}\cdot \frac92 = 17$.

We could have avoided fractions entirely by multiplying both sides of $\frac{2r-7}{9} = 3$ by 9 on the first step to get $9\cdot\frac{2r-7}{9} = 27$. Since \[9\cdot\frac{2r-7}{9} = \frac{9(2r-7)}{9} = \frac99(2r-7) = 2r-7,\]the 9’s cancel on the left side of $9\cdot\frac{2r-7}{9} = 27$ to leave $2r-7=27$. Adding 7 to both sides gives $2r = 34$, so $r=17$, as before.

Checking our answer, we find that if $r=17$, then $\frac{2r-7}{9} = \frac{2\cdot 17-7}{9} = \frac{27}{9} = 3$, as required.

(d) We start by getting rid of the fractions. We eliminate the denominator on the right by multiplying both sides by 7: \[7\cdot\dfrac{3x+4}{5} = 7\cdot\dfrac{2x-8}{7}.\]The 7’s on the right-hand side cancel, because \[7\cdot \dfrac{2x-8}{7} = \frac{7\cdot(2x-8)}{7} = \frac{7}{7} \cdot\frac{2x-8}{1} = 2x-8.\]So, we can write $7\cdot\frac{3x+4}{5} = 7\cdot\frac{2x-8}{7}$ as \[\frac{7(3x+4)}{5} = 2x - 8.\]Next, we multiply both sides by 5 to cancel the 5 in the denominator on the left-hand side: \[5\cdot \frac{7(3x+4)}{5}=5(2x-8).\]The 5’s on the left cancel, and we are left with \[ 7(3x+4) = 5(2x-8).\]Expanding both sides gives \[7(3x) + 7(4)=5(2x) - 5(8).\]Simplifying both sides gives $21x + 28=10x - 40 $, and now we’re in familiar territory. Subtracting $10x$ from both sides gives $11x + 28 = -40$. Subtracting 28 from both sides gives $11x = -68$. Dividing both sides by 11 gives $x = -\frac{68}{11}$.

Notice that multiplying both sides of $$ \dfrac{3x+4}{5}=\dfrac{2x-8}{7}$$by the denominators of both fractions gave us \[ 7(3x+4) = 5(2x-8).\]Rather than performing these multiplications as two separate steps, we will often perform both at once. Multiplying both sides of the original equation by 5 and by 7 gives \[5\cdot 7 \cdot \dfrac{3x+4}{5}=5\cdot 7\cdot \dfrac{2x-8}{7}.\]The 5 on the left cancels with the 5 in the denominator on the left, and the 7 on the right cancels with the 7 in the denominator on the right, leaving \[ 7(3x+4) = 5(2x-8).\]We call this process cross-multiplying.

Our last example above showed another way to simplify working with equations:

l
Concept:

If you don’t like dealing with fractions, you can eliminate fractions from a linear equation by multiplying both sides of the equation by a constant that cancels the denominators of the fractions.

Let’s practice this strategy.

Problem 5.15
t
V

Solve the following equations:

(a) $\color[rgb]{1,1,1} \dfrac95 -\dfrac{2x}{3} = \dfrac{6x}{5} + \dfrac73$

(b) $\color[rgb]{1,1,1}\dfrac{4-7t}{6} = \dfrac{t}{8} + 2$

Solution for Problem 5.15:

(a) Let’s get rid of the fractions right away. We multiply both sides of the equation by 3 to cancel the denominators that are 3, and multiply by 5 to cancel the denominators that are 5. Therefore, we can take care of both at once by multiplying by $3\cdot 5=15$. Using the distributive property to expand, the left-hand side becomes \begin{align*} 15\left(\frac95 -\frac{2x}{3}\right) &= 15\cdot \frac95 - 15\cdot\frac{2x}{3}\\ &=\frac{15}{5}\cdot9 -\frac{15}{3}\cdot 2x\\ &=27 - 5\cdot 2x\\ &= 27 - 10x. \end{align*}

Multiplying the right-hand side of the original equation by 15 gives \begin{align*} 15\left(\frac{6x}{5} + \frac73\right) &=15\cdot\frac{6x}{5} + 15\cdot \frac{7}{3}\\ &=\frac{15}{5}\cdot6x +\frac{15}{3}\cdot 7\\ &= 3\cdot 6x + 5\cdot 7\\ &=18x + 35. \end{align*}Combining this with our simplified left-hand side gives \[27-10x = 18x + 35.\]We add $10x$ to both sides to get $27=28x + 35$. We subtract 35 from both sides to get $-8 = 28x$ and divide by 28 to find $x=-\frac{8}{28} = -\frac{2}{7}$.

(b) We might start by multiplying both sides by $6\cdot 8$ to cancel both denominators. However, since $\text{lcm}[6,8]=24$, we can cancel both denominators by multiplying both sides by 24 instead of 48: \[24\left(\dfrac{4-7t}{6}\right) = 24\left(\frac{t}{8} + 2\right).\]Multiplying on the left-hand side and distributing on the right gives \[\frac{24(4-7t)}{6} = 24\cdot \frac{t}{8} + 24\cdot 2,\]so \[\frac{24}{6}(4-7t) = \frac{24}{8}t + 48.\]Dividing gives $4(4-7t) = 3t + 48$. No more fractions! Expanding the left-hand side gives us $16-28t = 3t + 48$. Adding $28t$ to both sides and subtracting 48 from both sides gives $-32 = 31t$. Dividing by 31 gives us $t = -\frac{32}{31}$.

So far, all the equations we have solved have had exactly one solution. This isn’t always the case!

Problem 5.16
t
V

(a) Find all values of $\color[rgb]{1,1,1}w$ that satisfy $\color[rgb]{1,1,1}5w + 3- 2w = w - 8 + 2w - 3$.

(b) Find all values of $\color[rgb]{1,1,1}z$ that satisfy $\color[rgb]{1,1,1}2z - 8 - 5z = 2 - 3z -10$.

Solution for Problem 5.16:

(a) We first simplify both sides. This gives us \[3w + 3 = 3w -11.\]When we next try to get all the $w$ terms on one side by subtracting $3w$ from both sides, we have \[3 = -11.\]

Uh-oh! What happened to the $w$s? They all canceled. Worse yet, we are left with an equation that can clearly never be true, since $3$ cannot ever equal $-11$!

Since the equation $3 = -11$ can never be true, we know that the original equation can never be true either. That is, the original equation is not true for any value of $w$. We can see why when we look back to the equation $3w + 3 = 3w-11$. The left-hand side is 14 greater than the right-hand side, no matter what value of $w$ we use.

We conclude that there are no solutions to the original equation.

(b) Once again, we simplify both sides of the equation, which gives \[-3z-8 = -3z -8.\]Since both sides of the equation simplify to the same expression, we see that the equation is always true! No matter what value of $z$ we choose, the equation will always be true. Therefore, all values of $z$ satisfy the given equation.

We see now that some linear equations have no solutions, and others that are satisfied by every value of the variable in the equation.

Z
Important:

If a linear equation can be manipulated into an equation that is never true (such as $\color[rgb]{0.11,0.21,0.37}3 = -11$), then there are no solutions to the equation.

If the two sides of an equation are equivalent, such as in the equation $\color[rgb]{0.11,0.21,0.37}-3z-8=-3z-8$, then all possible values of the variable are solutions to the original equation. Similarly, if a linear equation can be manipulated into an equation in which both sides are identical, then all possible values of the variable are solutions to the original equation. (The one exception to this is if one of the manipulations is multiplying both sides by 0, which is a pretty silly thing to do to a linear equation!)

Problem 5.17
t
V

For what value of $\color[rgb]{1,1,1}c$ do the equations $\color[rgb]{1,1,1}2y - 5 = 17$ and $\color[rgb]{1,1,1}cy - 8 = 36$ have the same solution for $\color[rgb]{1,1,1}y$?

Solution for Problem 5.17: We know how to handle the first equation, so let’s start there. By solving the first equation for $y$, we can find the value of $y$ that must satisfy both equations. Adding 5 to both sides of $2y - 5 = 17$ gives $2y = 22$. Dividing by 2 then gives $y=11$. This value of $y$ must also satisfy $cy-8=36$. So, when we substitute $y=11$ into $cy-8=36$, we must have a true equation. This substitution gives \[11c - 8=36.\]Now that we have a linear equation for $c$, we can find $c$. Adding 8 to both sides gives $11c = 44$. Dividing by $11$ then gives $c=4$.

Exercises
5.3.1:
t
V

Solve the following equations:

(a) $\color[rgb]{0.35,0.35,0.35}2x + 5 = 11$

Show Solution

(b) $\color[rgb]{0.35,0.35,0.35}\dfrac13 = -1\dfrac12- 6a$

Show Solution

(c) $\color[rgb]{0.35,0.35,0.35}-7t + 19 = 61$

Show Solution
5.3.2:
t
V

Solve the following equations:

(a) $\color[rgb]{0.35,0.35,0.35}3y + 9 = 2y + 1$

Show Solution

(b) $\color[rgb]{0.35,0.35,0.35}5x - 3 - x = 14 - 3x + 11$

Show Solution

(c) $\color[rgb]{0.35,0.35,0.35}1000a + 218 = 998a + 232$

Show Solution
5.3.3:
t
V

If $\color[rgb]{0.35,0.35,0.35}3x-2 = 11$, then what is the value of $\color[rgb]{0.35,0.35,0.35}6x + 5$?

Show Solution
5.3.4:
t
V

Solve the following equations:

(a) $\color[rgb]{0.35,0.35,0.35}\dfrac23t + \dfrac45 = -\dfrac12$

Show Solution

(b) $\color[rgb]{0.35,0.35,0.35}\dfrac12(z+3) = \dfrac13(z-7)$

Show Solution

(c) $\color[rgb]{0.35,0.35,0.35}\dfrac{4x}{7} - \dfrac12 = -\dfrac34 - \dfrac{2x}{5}$

Show Solution
5.3.5:
t
V

Solve $\color[rgb]{0.35,0.35,0.35}\dfrac{2x+7}{5} = -\dfrac{1-3x}{8}$.

Show Solution
5.3.6:
t
V

Solve the following equations:

(a) \[\color[rgb]{0.35,0.35,0.35} 2(z+3) - 5(6-z) = 8(3z+3)- 4(1 -2z) \]

Show Solution

(b) $\color[rgb]{0.35,0.35,0.35}\dfrac{m+11}{3} + \dfrac{m-2}{6} = \dfrac{2m-1}{12}$

Show Solution

(c) $\color[rgb]{0.35,0.35,0.35}\dfrac{p-2}{4} = \dfrac{2p-3}{8}$

Show Solution
a