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3 years ago

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Each course at college X is worth either 2 or 3 credits. The members of the men's swim team are taking a total of 48 courses tha

t are worth a total of 107 credits. How many 2-credit courses and how many 3-credit courses are being taken?

1 answer:

Serga [27]3 years ago

7 0

Answer:

Let the number of courses that are worth 3 credits each be x and those worth 4 credits be y. With the given information, you can write the following equations:

x + y = 48

3x + 4y = 155

You can solve the above equations by method of elimination/substitution

x + y = 48 ⇒ x = 48 - y (Now, substitution this equation into 3x + 4y = 155)

3(48 - y) + 4y = 155

144 -3y + 4y = 155

y + 144 = 155

y = 11

Now plug this solution back into x = 48 - y

x = 48 - 11 = 37

Check work (by plugging the solutions back into the 3x + 4y and see if it's equal to 155):

3(37) + 4(11) = 155

Answer: There are 37 of the 3-credit course and 11 of the 4-credit course

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3 years ago

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2/7m-1/7=3/14 solve step by step

yulyashka [42]

Solution for $\frac{2}{7}m - \frac{1}{7} = \frac{3}{14} \ is \ m = \frac{5}{4} \ or \ m = 1\frac{1}{4} \ or \ m = 1.25$

<h3>Further explanation</h3>

It is a case about one variable linear quations and we have to solve the equation to get the variable m. There are two ways to solve it!

Our main plan is to isolate the variable m alone at the end of the process on one side of the equation until the variable will be equal to the value on the opposite side.

<u>First way</u>

$\frac{2}{7}m - \frac{1}{7} = \frac{3}{14}$

Let us add $\frac{1}{7}$ to both sides

$\frac{2}{7}m - \frac{1}{7} + \frac{1}{7} = \frac{3}{14} + \frac{1}{7}$

$\frac{2}{7}m = \frac{3}{14} + \frac{1}{7}$

On the right side for the addition operation, we equate the common denominator by multiplying $\ \frac{1}{7} \ by \ \frac{2}{2}$

$\frac{2}{7}m = \frac{3}{14} + \frac{2}{14}$

Then we combine terms to get

$\frac{2}{7}m = \frac{5}{14}$

Divide by the coefficient of m, or in other words, multiply both sides by $\frac{7}{2}$

$\frac{2}{7}m \times \frac{7}{2} = \frac{5}{14} \times \frac{7}{2}$

Finally, the solution is obtained as follows

$m = \frac{35}{28}$

Simplify fractions, both the numerator and denominator are divided equally by 7.

$\boxed{ \ m = \frac{5}{4} \ }$

Convert into mixed fractions, we get:

$\boxed{ \ m = 1 \frac{1}{4} \ }$

In decimal form, we get

$m = 1 \frac{25}{100} \rightarrow \boxed{ \ m = 1.25 \ }$

<u>Second way (a quick way)</u>

$\frac{2}{7}m - \frac{1}{7} = \frac{3}{14}$

Because the denominators 7 and 14 have LCM = 14, both sides are multiplied by 14 (LCM is the Least Common Multiple)

$14 \times \big( \frac{2}{7}m - \frac{1}{7} \big) = 14\times \big( \frac{3}{14} \big)$

We use the distributive property of multiplication on the left side

$\frac{28}{7}m - \frac{14}{7} = \frac{42}{14}$

or it can also directly lead to

$4m - 2 = 3$

Add 2 to both sides, we get

$4m = 5$

Divide by the coefficient of m, or in other words, both sides are divided by 4

$\boxed{ \ m = \frac{5}{4} \ }$

Or, $\boxed{ \ m = 1 \frac{1}{4} \ }$

Or, $m = 1 \frac{25}{100} \rightarrow \boxed{ \ m = 1.25 \ }$

Wanna check the solution into the equation?

$\big( \frac{2}{7} \times \frac{5}{4} \big) - \frac{1}{7} = \frac{3}{14}$

$\frac{10}{28} - \frac{1}{7} = \frac{3}{14}$

$\frac{5}{14} - \frac{2}{14} = \frac{3}{14}$

$\frac{3}{14} = \frac{3}{14}$

Both sides show the same value, so the solution is correct.

Note:

Remember, how to manipulate both sides of the equation with the algebraic properties of equality such as:

adding,
subtracting,
multiplying, and/or
dividing both sides of the equation with the same number.

In the form of fractions, the steps that must be considered are

equate the denominator,
simplify fractions, and
for the final answer, convert fractions to mixed fractions or decimal forms

All these processes can occur repeatedly until the isolated variables are obtained on one side of the equation.

<h3>Learn more</h3>

A word problem that forms a single variable linear equation brainly.com/question/1566971
Learn more about the single variable linear equation that has no solution, has one solution, and has infinitely many solutions brainly.com/question/2595790
Questioning the stages of solving a word problem about one variable linear equations brainly.com/question/2038876

<h3>Answer details </h3>

Grade : Middle School

Subject : Mathematics

Chapter : Linear Equation in One Variable

Keywords : solve, solution, variable, coefficient, 2/7m - 1/7 = 3/14, 5/4, 1 1/4, 1.125, algebraic properties of equality, one, linear equation, isolated, manipulate, operations, add, subtract, multiply, divide, fraction, equate, denominator, numerator, both sides, decimal

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3 years ago

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