to view answer.Often in math problems and in real-life mathematical situations, we have an unknown that we need to “solve for”. When asked to solve for a variable in an equation, we are simply looking for the values of that variable that make the equation true.
Sometimes, we come create or come across an equation in which we have to find an unknown that is in the denominator of a fraction. For instance:
Find all possible values of x such that the statement is true:
$${\Large{1 \over 2} = {3 \over x}}$$
In this case, we can likely find the value of x simply by examining the problem.
You may think to yourself: “I want a fraction that is equal to $${\Large{1 \over 2}}$$ that has 3 on the top. That means that the proportion of 1 to 2 must be the same as the proportion of 3 to x. Since 1 is half of 2, then 3 must be half of x. Thus, the x must be 6”.
You may also solve the problem by multiplying each side of the equation by x. Since the x on the right side would cancel, this would give us:
$${\Large{1 \over 2}}x = 3$$
Now, you know once again that ½ of x is 3, so x must be 6. (Alternatively, you could multiply both sides by 2 to get the same result).
Now for a more difficult example:
$${\Large{3 \over {x + 2}} = {7 \over {5x - 4}}}$$
This problem is not so easily solved by examination. So, here’s how we can approach it:
1. For any math problem when we are trying to find the values of x that make the equation true, the first thing we should do is see if there are any values of x that will cause a serious problem (or contradiction). For instance, in this case, we are dividing by values added to x. Thus, we need to be careful that we are not using some value of x that will make the denominator zero. Which values of x will do this?
a. In this case, we don’t want either denominator to equal zero. So, we do not want: $$ x + 2 $$ or $$ 5x - 4 $$ to equal zero.
b. When does $$ x + 2 = 0 $$ ? When $$ x = - 2 $$
When does $$ 5x - 4 = 0 $$
c. So, when we find our final answer by manipulating the equation, we need to make sure that it is neither -2 nor $${\Large{4 \over 5}}$$, because these would give us a zero in the denominator.
Why can’t we divide by zero? You have likely heard in the past that if you divide by zero, your answer is “undefined”. What does this mean? Division can be rewritten as multiplication. For instance: $${\Large{{10} \over 2}} = 5 $$ means that $$ 5 \cdot 2 = 10 $$
$${\Large{{56} \over 7}} = 8 $$ means that $$ 7 \cdot 8 = 56 $$
Now, suppose that division by zero was defined. In other words, when we divide by zero, we will get an answer. So: Thus, be sure that you do not divide by zero. Your answer will not make sense mathematically. |
2. Now that we have found all of the values of x that absolutely cannot work, we can start trying to find the ones that will. First, we would like to get all of the x’s out of the denominators, since right now, they are part of the terms dividing 3 and 7. So, multiply both sides of the equation by $$ \left( {x + 2} \right) $$ :
$$
\left( {x + 2} \right) \cdot {\Large{3 \over {x + 2}}} = {\Large{7 \over {5x - 4}}} \cdot \left( {x + 2} \right)
$$
On the left side, $${\Large{{x + 2} \over {x + 2}}} = 1 $$ , so we are left with $$ 1 \cdot 3 $$ or just 3 on that side:
$$
3 = {\Large{7 \over {5x - 4}}} \cdot \left( {x + 2} \right)
$$
3. Now, we multiply both sides of the equation by the other denominator, $$ \left( {5x - 4} \right) $$
$$
\left( {5x - 4} \right) \cdot 3 = \left( {5x - 4} \right) \cdot {\Large{7 \over {5x - 4}}} \cdot \left( {x + 2} \right)
$$
Now, on the right side, we have $${\Large{{5x - 4} \over {5x - 4}}} = 1 $$ , so we are left with $$ 7\left( {x + 2} \right) on that side:
$$
\left( {5x - 4} \right) \cdot 3 = 7 \cdot \left( {x + 2} \right)
$$
4. Now, use the distributive property on both sets of parentheses, to get:
$$
15x - 12 = 7x + 14
$$
5. Finally, solve this linear equation by getting x alone on one side:
Add 12 to each side:
$$
15x - 12 + 12 = 7x + 14 + 12
$$
$$
15x = 7x + 26
$$
Subtract 7x from each side:
$$
15x - 7x = 7x - 7x + 26
$$
$$
8x = 26
$$
Divide each side by 8:
$${\Large{{8x} \over 8} = {{26} \over 8}}$$
$$
x = {\Large{{26} \over 8}}$$
Simplify the fraction:
$$
x = {\Large{{26} \over 8} = {{13} \over 4}}$$
6. The very last thing we need to do is make sure that our answer is not one of the values of x that we determined in step 1 to be a problem value. Our answer is neither -2 nor $${\Large{4 \over 5}}$$ . So, we are done.
Some practice problems to check your skills:
1. $${\Large{{12} \over {13}} = {7 \over x}}$$
Think about it for a moment and then access this link to view answer.
2. $${\Large{3 \over 5} = {2 \over {x + 3}}}$$
Think about it for a moment and then access this link to view answer.
3. $${\Large{7 \over {9 - y}} = {5 \over 7}}$$
Think about it for a moment and then access this link to view answer.
4. $${\Large{3 \over {2 - x}} = {7 \over {x + 1}}}$$
Think about it for a moment and then access this link to view answer.
5. $${\Large{{23} \over {4x - 2}} = {{15} \over {7x + 4}}}$$
Think about it for a moment and then access this link to view answer.