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Subtraction is used to undo an addition.
Addition is used to undo a subtraction.
The procedure is illustrated in the problems of [link] .
Use the addition/subtraction property of equality to solve each equation.
$x+4=6$ .
4 is associated with
$x$ by addition. Undo the association by
subtracting 4 from
both sides.
$\begin{array}{c}x+4-4=6-4\hfill \\ x+0=2\hfill \\ x=2\hfill \end{array}$
Check: When $x=2$ , $x+4$ becomes
The solution to $x+4=6$ is $x=2$ .
$m-8=5$ . 8 is associated with $m$ by subtraction. Undo the association by adding 8 to both sides.
$\begin{array}{c}m-8+8=5+8\hfill \\ m+0=\text{13}\hfill \\ m=\text{13}\hfill \end{array}$
Check: When $m=\text{13}$ ,
becomes
a true statement.
The solution to $m-8=5$ is $m=\text{13}$ .
$-3-5=y-2+8$ . Before we use the addition/subtraction property, we should simplify as much as possible.
$-3-5=y-2+8$
$-8=y+6$
6 is associated with
$y$ by addition. Undo the association by
subtracting 6 from
both sides.
$\begin{array}{c}-8-6=y+6-6\hfill \\ -\text{14}=y+0\hfill \\ -\text{14}=y\hfill \end{array}$
This is equivalent to
$y=-\text{14}$ .
Check: When $y=-\text{14}$ ,
$-3-5=y-2+8$
becomes
,
a true statement.
The solution to $-3-5=y-2+8$ is $y=-\text{14}$ .
$-\mathrm{5a}+1+\mathrm{6a}=-2$ . Begin by simplifying the left side of the equation.
$\underset{-5+6=1}{\underbrace{-5a+1+6a}}=-2$
$a+1=-2$ 1 is associated with $a$ by addition. Undo the association by subtracting 1 from both sides.
$\begin{array}{c}a+1-1=-2-1\hfill \\ a+0=-3\hfill \\ a=-3\hfill \end{array}$
Check: When $a=-3$ ,
$-\mathrm{5a}+1+\mathrm{6a}=-2$
becomes
,
a true statement.
The solution to $-5a+1+6a=-2$ is $a=-3$ .
$7k-4=6k+1$ . In this equation, the variable appears on both sides. We need to isolate it on one side. Although we can choose either side, it will be more convenient to choose the side with the larger coefficient. Since 8 is greater than 6, we’ll isolate $k$ on the left side.
$7k-4=6k+1$ Since $6k$ represents $+6k$ , subtract $6k$ from each side.
$\underset{7-6=1}{\underbrace{7k-4-6k}}=\underset{6-6=0}{\underbrace{6k+1-6k}}$
$k-4=1$ 4 is associated with $k$ by subtraction. Undo the association by adding 4 to both sides.
$\begin{array}{c}k-4+4=1+4\hfill \\ k=5\hfill \end{array}$
Check: When $k=5$ ,
$7k-4=6k+1$
becomes
a true statement.
The solution to $7k-4=6k+1$ is $k=5$ .
$-8+x=5$ . -8 is associated with $x$ by addition. Undo the by subtracting -8 from both sides. Subtracting -8 we get $-\left(-8\right)\text{=+}8$ . We actually add 8 to both sides.
$-8+x+8=5+8$
$x=\text{13}$
Check: When $x=\text{13}$
$-8+x=5$
becomes
,
a true statement.
The solution to $-8+x=5$ is $x=\text{13}$ .
For the following 10 problems, verify that each given value is a solution to the given equation.
$x-\text{11}=5$ , $x=\text{16}$
Substitute
$x=4$ into the equation
$4x-\text{11}=5$ .
$\begin{array}{}\text{16}-\text{11}=5\\ 5=5\end{array}$
$x=4$ is a solution.
$y-4=-6$ , $y=-2$
$2m-1=1$ , $m=1$
Substitute
$m=1$ into the equation
$2m-1=1$ .
$m=1$ is a solution.
$5y+6=-\text{14}$ , $y=-4$
$\mathrm{3x}+2-\mathrm{7x}=-\mathrm{5x}-6$ , $x=-8$
Substitute
$x=-8$ into the equation
$\mathrm{3x}+2-7=-\mathrm{5x}-6$ .
$x=-8$ is a solution.
$-6a+3+3a=4a+7-3a$ , $a=-1$
$-8+x=-8$ , $x=0$
Substitute
$x=0$ into the equation
$-8+x=-8$ .
$x=0$ is a solution.
$8b+6=6-5b$ , $b=0$
$4x-5=6x-\text{20}$ , $x=\frac{\text{15}}{2}$
Substitute
$x=\frac{\text{15}}{2}$ into the equation
$4x-5=6x-\text{20}$ .
$x=\frac{\text{15}}{2}$ is a solution.
$-3y+7=2y-\text{15}$ , $y=\frac{\text{22}}{5}$
Solve each equation. Be sure to check each result.
$y-8+\text{10}=2$
$y-2+5=4$
$z+\text{10}-8=-8+\text{10}$
$-5+3=h-4$
$5a+6=4a-8$
$\text{12}h-1-3-5h=2h+5h+3(-4)$
$-9n-2-6+5n=3n-\left(2\right)\left(-5\right)-6n$
$y-2\text{.}\text{161}=5\text{.}\text{063}$
$y=7\text{.}\text{224}$
$a-\text{44}\text{.}\text{0014}=-\text{21}\text{.}\text{1625}$
$-0\text{.}\text{362}-0\text{.}\text{416}=5\text{.}\text{63}m-4\text{.}\text{63}m$
$m=-0\text{.}\text{778}$
$8\text{.}\text{078}-9\text{.}\text{112}=2\text{.}\text{106}y-1\text{.}\text{106}y$
$4\text{.}\text{23}k+3\text{.}\text{18}=3\text{.}\text{23}k-5\text{.}\text{83}$
$k=-9\text{.}\text{01}$
$6\text{.}\text{1185}x-4\text{.}\text{0031}=5\text{.}\text{1185}x-0\text{.}\text{0058}$
$\text{21}\text{.}\text{63}y+\text{12}\text{.}\text{40}-5\text{.}\text{09}y=6\text{.}\text{11}y-\text{15}\text{.}\text{66}+9\text{.}\text{43}y$
$y=-\text{28}\text{.}\text{06}$
$0\text{.}\text{029}a-0\text{.}\text{013}-0\text{.}\text{034}-0\text{.}\text{057}=-0\text{.}\text{038}+0\text{.}\text{56}+1\text{.}\text{01}a$
( [link] ) Is $\frac{7\text{calculators}}{\text{12}\text{students}}$ an example of a ratio or a rate?
rate
( [link] ) Convert $\frac{3}{8}\text{}$ % to a decimal.
( [link] ) Use the clustering method to estimate the sum: $\text{89}+\text{93}+\text{206}+\text{198}+\text{91}$
( [link] ) Combine like terms: $4x+8y+\text{12}y+9x-2y$ .
$\text{13}x+\text{18}y$
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