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Iteru [2.4K]
3 years ago
12

there are 5 more than twice as many students taking Algebra 1 then taking algebra 2. If there are 44 students taking Algebra 2,

what is the least number of students who could be taking Algebra 1? The least number of students is?
Mathematics
2 answers:
Temka [501]3 years ago
8 0

Answer:

93

Step-by-step explanation:

Key :

A1 = Algebra 1

A2 = Algebra 2

Alright so basically lets first look at the info they gave us :

We have 5 more than twice as many students taking A1 than we do A2.

We have 44 students taking A2.

And we need to find the least amount of students that could be taking A1.

So we need to take the amount of students taking A2 (44) and double it to find the amount taking A1.

So we can do 44 x 2 = 88 to get this.

But the problem also states there is 5 more then twice the number of students taking A2.

So we have that 88 but now we just need to add 5 to make up for them telling us that in the problem.

So :

88 + 5 = 93

Our final answer and least amount of students taking A1 is 93 students.

lakkis [162]3 years ago
5 0

Answer:

93 students

Step-by-step explanation:

44 x 2 = 88

88 + 5 = 93

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The <u>correct answer</u> is:

D) \left \{ {{2x-y=7} \atop {2x+7y=31}} \right..

Explanation:

We solve each system to find the correct answer.

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\left \{ {{3x-2y=9} \atop {3x+2y=14}} \right.

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Since the coefficients of y are -2 and 2, we can add the equations to solve, since -2+2=0 and cancels the y variable:
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<u>For B</u>:
\left \{ {{x-y=-2} \atop {4x-3y=11}} \right.

For this equation, it will be easier to isolate a variable and use <u>substitution</u>, since the coefficient of both x and y in the first equation is 1:
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Combining like terms, we have:
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x+2y=-11

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2x-y=7
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Add 3 to each side:
2x-3+3=7+3
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Divide each side by 2:
2x/2=10/2
x=5

This gives us the x- and y-coordinate we need, so <u>D is the correct answer</u>.
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