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lbvjy [14]
3 years ago
15

How would i solve for x and y in 3x+2y=17 and 2x-y=2 by using substitution

Mathematics
1 answer:
maxonik [38]3 years ago
4 0
I’ll do an example problem, and I challenge you to do this on your own!
4x+6y=23
7y-8x=5
Solving for y in 4x+6y=23, we can separate the y by subtracting both sides by 4x (addition property of equality), resulting in 6y=23-4x. To make the y separate from everything else, we divide by 6, resulting in (23-4x)/6=y. To solve for x, we can do something similar - subtract 6y from both sides to get 23-6y=4x. Next, divide both sides by 4 to get (23-6y)/4=x.

Since we know that (23-4x)/6=y, we can plug that into 7y-8x=5, resulting in
7*(23-4x)/6-8x=5
= (161-28x)/6-8x
Multiplying both sides by 6, we get 161-28x-48x=30
= 161-76x
Subtracting 161 from both sides, we get -131=-76x. Next, we can divide both sides by -76 to separate the x and get x=131/76. Plugging that into 4x+6y=23, we get 4(131/76)+6y=23. Subtracting 4(131/76) from both sides, we get
6y=23-524/76. Lastly, we can divide both sides by 6 to get y=(23-524/76)/6

Good luck, and feel free to ask any questions!
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posledela
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Hope this helps :)
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3 years ago
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EleoNora [17]

Answer:

The probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

Step-by-step explanation:

Let the random variable <em>X</em> represent the miles-per-gallon rating of passenger cars.

It is provided that X\sim N(\mu=33.8,\ \sigma^{2}=3.5^{2}).

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P(X>37.3)=P(\frac{X-\mu}{\sigma}>\frac{37.3-33.8}{3.5})

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Thus, the probability that a randomly selected passenger car gets more than 37.3 mpg is 0.1587.

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2 years ago
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Answer:

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Step-by-step explanation:

1/6 + 4/6 = 5/6

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