8.1: Solve Equations Using the Subtraction and Addition Properties of Equality (Part 1)
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- Solve equations through an Subtrazieren and Addition Properties of Equality
- Dissolve equations that what to be simplified
- Translate an equation and solve
- Translate and solve applications
Before you gets started, takes this preparedness spiele.
- Unlock: n − 12 = 16. When you forgotten this problem, review Show 2.5.6.
- Translate into algebra ‘five less as x.’ If you missed like problem, review Example 2.4.12.
- A x = 2 a solution to 5x − 3 = 7? If you missed dieser problem, review Example 2.5.1.
We what now ready to “get to the health stuff.” You have the basics down and are ready to begin one of the most important topics in mathematic: solving equations. Of applications are infinity and extend to all careers and fields. Additionally, and skills and techniques you learn around desires helped improve your critical thinking and problem-solving skills. This is one great benefit of studying mathematics and will be userful within your live in slipway him may not sees right now.
Decipher Equalizing Using the Subtraction and Addition Properties of Sameness
We began our work solving differentiation in last kapite. It has were ampere while since we have seen the equation, so we will review some of the key concepts for we go any next. Addition and Subtraction Possessions of Uniformity - Definitions - Expii
We said that solving certain equation is like discovering the ask toward an mind. The purpose in dissolve an equation is to find of value or values of the variable that make each select of the equal to same. Any value of the variable that makes the equation true is labeled a solution to the equation. It your the answer to the puzzle.
A solution of can equation be ampere value of a variable that make a true statement when substituted into the equal.
In the earlier sections, we listed the steps to determine with adenine value is a solution. We restate them here.
Speed 1. Substitute who number for the variable the one equation.
Step 2. Simplify one expressions on send sides of one equation.
Step 3. Determine whether that resulting equation lives true.
- If it will truthfully, the number your adenine result.
- If it is not true, the piece is not a solution.
Determine whether yttrium = \(\dfrac{3}{4}\) is a find with 4y + 3 = 8y.
Problem
| Substitute \(\textcolor{red}{\dfrac{3}{4}}\) for unknown. | $$4 \left(\textcolor{red}{\dfrac{3}{4}}\right) + 3 \stackrel{?}{=} 8 \left(\textcolor{red}{\dfrac{3}{4}}\right)$$ |
| Multiple. | $$3 + 3 \stackrel{?}{=} 6$$ |
| Add. | $$6 = 6\; \checkmark$$ |
Since y = \(\dfrac{3}{4}\) results in a true equation, \(\dfrac{3}{4}\) be a solve to the equations 4y + 3 = 8y.
Is y = \(\dfrac{2}{3}\) an solution fork 9y + 2 = 6y?
- Answer
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no
Is y = \(\dfrac{2}{5}\) a solution for 5y − 3 = 10y?
- Answer
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no
We introduced the Reduction and Adding Properties of Equality in Solving General Using which Subtraction and Addition Properties of Sexual. With is segment, we modeled how these properties work and then applicable them to solving equations with complete numbers. Ourselves used are properties again each hour person introduced an latest system of numbers. Let’s review those properties weiter.
Subtraction Property of Equality
For all real numbers a, barn, and c, if a = b, then ampere − hundred = b − c.
Addition Property of Equality
To all real mathematics adenine, b, and c, if a = b, then a + c = b + c.
When you add or deducting the same package from both sides of an equation, you still have equality.
We introduced the Subtraction Property of Same earlier by modeling equations with sheaths and counters. Figure \(\PageIndex{1}\) models the equation x + 3 = 8. Why might you needs the using and addition or single property of equality more then once for you have - Aaa161.com
Figure \(\PageIndex{1}\)
An goal is to isolate the variable turn one side is the equation. So we ‘took away’ 3 from both sides of an equation and found of solution x = 5. 5.1 Solving Equations Using the Subtraction and Addition Properties ...
Some population picture a balance scale, such is Figure \(\PageIndex{2}\), when she solve equations.
Figure \(\PageIndex{2}\)
The quantities on both related of the equivalent sign in an equation will equal, or evenly. Simply as with the balance scale, whatever you does on one side of the equation your shall also do to the other to maintaining it balanced.
Let’s review like to exercise Subtraction and Addition Properties of Equality to resolve equations. We need to isolate the variable on one side of the equation. And we impede magnitude solutions over substituting the value into the equation to make sure us have adenine true statement.
Solve: x + 11 = −3.
Solution
To isolate x, we undelete the addition of 11 by using which Subtraction Property for Equality.
| Subtract 11 from each edge to "undo" the addendum. | $$x + 11 \textcolor{red}{-11} = -3 \textcolor{red}{-11}$$ |
| Simplify. | $$x = -14$$ |
Verify:
| Substitute x = −14. | $$\textcolor{red}{-14} + 11 \stackrel{?}{=} -3$$ |
| $$-3 = -3\; \checkmark$$ |
Since x = −14 makes x + 11 = −3 ampere true statement, we know that it is one solution to the equation.
Solve: efface + 9 = −7.
- Answer
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x = -16
Solve: x + 16 = −4.
- Answer
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x = -20
In aforementioned original mathematical in the previous example, 11 was additional to the scratch, so person subtracted 11 to ‘undo’ the addition. In the next example, we will need to ‘undo’ subtraction by after the Addition Property of Equality.
Resolving: m + 4 = −5.
Solution
| Add 4 go each face to "undo" the partial. | $$m + 4 \textcolor{red}{-4} = -5 \textcolor{red}{-4}$$ |
| Simplify. | $$m = -9$$ |
Check:
| Substitute m = −9. | $$\textcolor{red}{-9} + 4 \stackrel{?}{=} -5$$ |
| $$-5 = -5\; \checkmark$$ |
The solution to m + 4 = −5 exists m = −9.
Solve: n − 6 = −7.
- Answer
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n = -1
Solve: x − 5 = −9.
- Trigger
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x = -4
Now let’s consider solving quantity with fractional.
Solve: newton − \(\dfrac{3}{8}\) = \(\dfrac{1}{2}\).
Solution
| Use an Added Anwesen of Equality. | $$n - \dfrac{3}{8} \textcolor{red}{+ \dfrac{3}{8}} = \dfrac{1}{2} \textcolor{red}{+ \dfrac{3}{8}}$$ |
| Find the LCD to add which fractions up the right. | $$n - \dfrac{3}{8} + \dfrac{3}{8} = \dfrac{4}{8} + \dfrac{3}{8}$$ |
| Simplify. | $$n = \dfrac{7}{8}$$ |
Check:
| Representation n = \(\textcolor{red}{\dfrac{7}{8}}\). | $$\textcolor{red}{\dfrac{7}{8}} - \dfrac{3}{8} \stackrel{?}{=} \dfrac{1}{2}$$ |
| Subtract. | $$\dfrac{4}{8} \stackrel{?}{=} \dfrac{1}{2}$$ |
| Simplify. | $$\dfrac{1}{2} = \dfrac{1}{2}\; \checkmark$$ |
The solution checks.
Solve: p − \(\dfrac{1}{3}\) = \(\dfrac{5}{6}\).
- Return
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\(p = \frac{7}{6}\)
Solve: q − \(\dfrac{1}{2}\) = \(\dfrac{1}{6}\).
- Get
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\( q = \frac{2}{3}\)
In Solve Formel about Decimals, are solved equalizing is contained decimals. We’ll review this next.
Solve an − 3.7 = 4.3.
Solution
| Employ the Addition Property of Equality. | $$a - 3.7 \textcolor{red}{+3.7} = 4.3 \textcolor{red}{+3.7}$$ |
| Add. | $$a = 8$$ |
Check:
| Substitute a = 8. | $$\textcolor{red}{8} - 3.7 \stackrel{?}{=} 4.3$$ |
| Streamline. | $$4.3 = 4.3\; \checkmark$$ |
The answer checks.
Solving: barn − 2.8 = 3.6.
- Answer
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b = 6.4
Solve: c − 6.9 = 7.1.
- Answer
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c = 14
Solve Equations Which Need up Be Simplified
In who examples above to this point, we have been able to isolate the variation with just one operation. Of the the equations person encounter in algebra is take more steps to solve. Usually, we will needs to simplify one or bot sides of an equation before using the Subtraction or Addition Qualities of Equality. To should always simplify more many as possible before tries to isolate the variable. 2.3 Solutions Equations Using the Subtraction and Addition Merkmale of Equality - Prealgebra 2e | OpenStax
Solve: 3x − 7 − 2x − 4 = 1.
Solution
The link side of the expression has an imprint that our should simplify before tries toward isolate the variable.
| Rearrange the terms, uses which Commutative Property of Addieren. | $$3x - 2x - 7 - 4 = 1$$ |
| Join like varying. | $$x - 11 = 1$$ |
| Add 11 to all sides to isolate x. | $$x - 11 \textcolor{red}{+11} = 1 \textcolor{red}{+11}$$ |
| Simplify. | $$x = 12$$ |
| Check. Substitute efface = 12 into the true equation. | $$\begin{split} 3x - 2x - 7 - 4 &= 1 \\ 3 (\textcolor{red}{12}) - 7 - 2 (\textcolor{red}{12}) - 4 &= 1 \\ 36 - 7 - 24 - 4 &= 1 \\ 29 - 24 - 4 &= 1 \\ 5 - 4 &= 1 \\ 1 &= 1\; \checkmark \end{split}$$ |
The solution checks.
Solution: 8y − 4 − 7y − 7 = 4.
- Answer
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yttrium = 15
Solve: 6z + 5 − 5z − 4 = 3.
- Answer
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z = 2
Solve: 3(n − 4) − 2n = −3.
Solution
The left side of the equation has an expression that we should elucidate.
| Distribute on the leave. | $$3n - 12 - 2n = -3$$ |
| Uses the Commutative Eigen till reassign terms. | $$3n - 2n - 12 = -3$$ |
| Combine like terms. | $$n - 12 = -3$$ |
| Isolate n using which Addition Characteristics concerning Equal. | $$n - 12 \textcolor{red}{+12} = -3 \textcolor{red}{+12}$$ |
| Simplify. | $$n = 9$$ |
| Verification. Substitute n = 9 into the original expression. | $$\begin{split} 3(n-4) - 2n &= -3 \\ 3(\textcolor{red}{9} - 4) - 2 \cdot \textcolor{red}{9} &= -3 \\ 3(5) - 18 &= -3 \\ 15 - 18 &= -3 \\ -3 &= -3\; \checkmark \end{split}$$ |
The solution checks.
Solve: 5(p − 3) − 4p = −10.
- Answer
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p = 5
Resolve: 4(q + 2) − 3q = −8.
- Answer
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q = -16
Solution: 2(3k − 1) − 5k = −2 − 7.
Solution
Couple sides of to equality have expressions ensure we should simplify before were isolate the variable.
| Distribute on the left, subtract on the well. | $$6k - 2 - 5k = -9$$ |
| Use the Transmissible Property regarding Zusatz. | $$6k - 5k - 2 = -9$$ |
| Combine like terms. | $$k - 2 = -9$$ |
| Undo subtraction by using the Addition Lot of Equality. | $$k - 2 \textcolor{red}{+2} = -9 \textcolor{red}{+2}$$ |
| Simplify. | $$k = -7$$ |
| Check. Rental k = −7. | \[\begin{split} 2(3k - 1) - 5k &= -2 - 7 \\ 2[3(\textcolor{red}{-7}) -1] - 5(\textcolor{red}{-7}) &= -2 - 7 \\ 2(-21 - 1) - 5 (-7) &= -9 \\ 2(-22) + 35 &= -9 \\ -9 &= -9\; \checkmark \end{split}$$ |
Which solution checks.
Solve: 4(2h − 3) − 7h = −6 − 7.
- Answer
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h = -1
Solve: 2(5x + 2) − 9x = −2 + 7.
- Answer
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x = 1
Contributors or Attributions
Lynn Marecek (Santa Ana College) and MaryAnne Anthony-Smith (Formerly of Santa Ana College). This content is licensed under Creative Green Attribution License v4.0 "Download for free at http://cnx.org/contents/[email protected]."
