Problem 27
Question
Solve each equation using the addition property of equality. Be sure to check your proposed solutions. $$x-\frac{3}{4}=\frac{9}{2}$$
Step-by-Step Solution
Verified Answer
The solution for the equation \(x - \frac{3}{4} = \frac{9}{2}\) is \(x = \frac{21}{4}\) or \(5.25\) in decimal.
1Step 1: Add \(\frac{3}{4}\) to both sides
Since we need to isolate \(x\), we add \(\frac{3}{4}\) on both sides of the equation. So, \(x - \frac{3}{4} + \frac{3}{4} = \frac{9}{2} + \frac{3}{4}\). This simplifies to \(x = \frac{9}{2} + \frac{3}{4}\).
2Step 2: Simplify the Resulting Equation
The next step is to convert these fractions into equivalent fractions with the same denominator. So, \(x = \frac{18}{4} + \frac{3}{4}\), simplifying this further gives us: \(x = \frac{21}{4}\).
3Step 3: Check Your Answer
Substitute \(x\) into the original equation to make sure the equality holds true: \(x - \frac{3}{4} = \frac{9}{2}\) becomes \(\frac{21}{4} - \frac{3}{4} = \frac{9}{2}\). After simplifying we get \(\frac{18}{4} = \frac{9}{2}\), thus our solution is correct as both sides are equal.
Key Concepts
Addition Property of EqualityEquivalent FractionsChecking Solutions in Algebra
Addition Property of Equality
The addition property of equality is a fundamental principle in algebra which states that when you add the same number to both sides of an equation, the two sides remain equal. It's like balancing scales; if you add the same weight to both sides, the balance is maintained. Let's see how this works with our example equation:
We start with an equation where the variable we're solving for, in this case, \(x\), has a number subtracted from it: \(x - \frac{3}{4} = \frac{9}{2}\). To solve for \(x\), we need to 'get rid of' the \(\frac{3}{4}\) on the left side. According to the addition property of equality, we can do so by adding \(\frac{3}{4}\) to both sides. This gives us \(x = \frac{9}{2} + \frac{3}{4}\) as the \(\frac{3}{4}\) on the left cancels out. This process is essential because it helps keep the equation balanced as we work towards isolating the variable.
We start with an equation where the variable we're solving for, in this case, \(x\), has a number subtracted from it: \(x - \frac{3}{4} = \frac{9}{2}\). To solve for \(x\), we need to 'get rid of' the \(\frac{3}{4}\) on the left side. According to the addition property of equality, we can do so by adding \(\frac{3}{4}\) to both sides. This gives us \(x = \frac{9}{2} + \frac{3}{4}\) as the \(\frac{3}{4}\) on the left cancels out. This process is essential because it helps keep the equation balanced as we work towards isolating the variable.
Equivalent Fractions
Equivalent fractions are different fractions that represent the same part of a whole. They’re created by multiplying or dividing both the numerator (top number) and the denominator (bottom number) by the same number. Understanding equivalent fractions is crucial when solving equations that involve fractions, as it allows us to combine them by finding a common denominator.
For instance, when we have \(\frac{9}{2} + \frac{3}{4}\), we need to find an equivalent fraction for \(\frac{9}{2}\) that has 4 as the denominator so that we can easily add it to \(\frac{3}{4}\). By converting \(\frac{9}{2}\) to \(\frac{18}{4}\), we’ve found an equivalent fraction to \(\frac{9}{2}\), and now we can add \(\frac{18}{4}\) and \(\frac{3}{4}\) together to get \(\frac{21}{4}\). Understanding how to find and use equivalent fractions is a vital tool in algebra that simplifies complex equations into more manageable ones.
For instance, when we have \(\frac{9}{2} + \frac{3}{4}\), we need to find an equivalent fraction for \(\frac{9}{2}\) that has 4 as the denominator so that we can easily add it to \(\frac{3}{4}\). By converting \(\frac{9}{2}\) to \(\frac{18}{4}\), we’ve found an equivalent fraction to \(\frac{9}{2}\), and now we can add \(\frac{18}{4}\) and \(\frac{3}{4}\) together to get \(\frac{21}{4}\). Understanding how to find and use equivalent fractions is a vital tool in algebra that simplifies complex equations into more manageable ones.
Checking Solutions in Algebra
Checking solutions is the final and an essential step in solving algebraic equations. Once you find a solution, you must ensure it makes the original equation true. To check the solution, simply substitute the variable you solved for with the value you found into the given equation and then simplify.
In our example, we’ve found that \(x = \frac{21}{4}\). To check whether this is correct, we substitute \(x\) with \(\frac{21}{4}\) back into the original equation: \(\frac{21}{4} - \frac{3}{4} = \frac{9}{2}\). After simplifying both sides, we get \(\frac{18}{4} = \frac{9}{2}\), confirming that our solution of \(\frac{21}{4}\) makes the equation true. Any time you solve an equation, make sure to do this check; if the equation balances out, you can be confident in your solution. It's like giving a 'proof' that your answer is indeed the right one, adding a level of assurance to your algebra skills.
In our example, we’ve found that \(x = \frac{21}{4}\). To check whether this is correct, we substitute \(x\) with \(\frac{21}{4}\) back into the original equation: \(\frac{21}{4} - \frac{3}{4} = \frac{9}{2}\). After simplifying both sides, we get \(\frac{18}{4} = \frac{9}{2}\), confirming that our solution of \(\frac{21}{4}\) makes the equation true. Any time you solve an equation, make sure to do this check; if the equation balances out, you can be confident in your solution. It's like giving a 'proof' that your answer is indeed the right one, adding a level of assurance to your algebra skills.
Other exercises in this chapter
Problem 26
Solve each equation in using the multiplication property of equality. Be sure to check your proposed $$2 x-12 x=50$$
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Solve each equation. Be sure to check your proposed solution by substituting it for the variable in the original equation. $$7(3 x-2)+5=6(2 x-1)+24$$
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Solve the formula for the volume of a circular cylinder for \(h\)
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Use the addition property of inequality to solve each inequality and graph the solution set on a number line. \(3 x+4 \leq 2 x+7\)
View solution