The next step Juan will take is to multiply the second fraction by (x-2)/(x-2).
Given that Juan needs to rewrite this difference as one expression (3x/(x²-7x+10))-(2x/(3x-15)) and factor the denominator as (3x/(x-2)(x-5))-(2x/3(x-5)).
An algebraic expression in mathematics is an expression composed of variables and constants and algebraic operations (addition, subtraction, etc.). Expressions are made up of concepts.
The given expression is (3x/(x²-7x+10))-(2x/(3x-15))
Firstly, we will factored the denominator as
(3x/(x-2)(x-5))-(2x/3(x-5)).
Now, we will multiply and divide the second fraction by (x-2), we get
(3x/(x-2)(x-5))-((2x(x-2))/3(x-5)(x-2)).
Hence, the next step when subtracting these expressions (3x/(x-2)(x-5))-(2x/3(x-5)) is multiply the second fraction by (x-2)/(x-2).
Learn more about algebraic expression from here brainly.com/question/4344214
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Step-by-step explanation:
Let "c" be the original number of classrooms.
The 1200/c was the original number of students per classroom.
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Equation:
1200/(c-4) = (1200/c)+10
Multiply thru by c(c-4)
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1200c = 1200(c-4)+10c(c-4)
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1200c = 1200c-4800 + 10c^2-40c
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10c^2-40c-4800 = 0
c^2-4c-480 = 0
Factor:
(c-24)(c+20) = 0
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Positive solution:
c = 24 (# of classrooms originaly planned)
...6 and 10 are roots, so x – 6 and x – 10 are factors.
y = a(x – 6)(x – 10).....plug in the point (8, 2) and solve for a:
2 = a(8 – 6)(8 – 10)
2 = –4a
a = –1/2
...y = (–1/2)(x – 6)(x – 10)
...y = (–1/2)(x² – 16x + 60)
...y = (–x²/2) + 8x – 30 <<<------Answer, or:
...y = (–1/2)(x – 8)² + 2 <<<------Answer
