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Lana71 [14]
3 years ago
13

Review your solution to part (b) of problem 5-13. You probably multiplied each fraction by the lowest (least) common denominator

(LCD), which, in this case, is . But what would you do if you could not determine the LCD? You will explore this idea in parts (a) through (e) below. Multiply the three denominators: . Multiply both sides of the equation by the product you got in part (a). This gives you an equivalent equation. Have your partner multiply both sides of the equation by 10 and find a different equivalent equation. Now, solve your equation for and check your answer. While you solve your equation, your partner will be solving his or her equation. Compare your result with your partner's. Why do both your method and your partner's method work? That is, why do both using the product of the denominators and using the LCD of both successfully eliminate fractions from equations?

Mathematics
1 answer:
EleoNora [17]3 years ago
7 0
<h3>Missing Part of Question:</h3>

Some information was missing so I am attaching the complete question.

<h3></h3><h3>Step-by-step explanation:</h3>

Given Equation is:

  \frac{3y}{5}\;+\;\frac{3}{2}\;=\;\frac{7y}{10}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(1)

Step (a)

Let, A = 5 * 3 * 10

      A = 150

Step (b)

Multiplying both sides of equation (1) with A = 150, we get,

90y_{_1}\; + \;225\; =\; 105y_{_1}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(a)

Multiplying both sides of equation (1) by 10 we get,

6y_{_2}\; + \;15\; =\; 7y_{_2}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;------\;\;(b)

Step (c)

Solving Equation (a), we get

105y_{_1}\;-\; 90y_{_1}\;=\;225\\\\15y_{_1}\;= \;225\\\\y_{_1}\;=\;\frac{225}{15}\\\\y_{_1}\;=\;15

Solving Equation (b), we get

7y_{_{2}}\; -\; 6y_{_{2}}\; = \;15\\\\y_{_{2}}\;=\; 15

Step (d)

y{_{_1}} \;= \;15; \;\;\;\;\;\&\;\;\;\;\;y{_{_2}}\; = \;15;\\\\\therefore y{_{_1}}\; =\; y{_{_2}}

Step (e)

Both methods work the same because as long as we multiply (or divide) the same amount to the total equation, the equation balances itself out. For example, consider an equation

x = 2;

If we multiply it by 10 on both sides we will get

10x = 20,

If we simplify the above equation, however, we will get the same initial equation.

x = 20/10

x = 2

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