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puteri [66]
3 years ago
14

Solve the system of linear equations by elimination −2x+3y=7 and 5x+8y=−2

Mathematics
1 answer:
forsale [732]3 years ago
4 0
You would need to make them eliminate so I would multiply the entire first one by 2.5 to make the X's cancel out. So you would end up with -5X+7.5Y=17.5. The X's would cancel and you would be left with 15.5Y=15.5. So Y=1. Then replace Y into one of the equations. I'm going to use the top again. -2X+3(1)=7. Subtract 3 from both sides to get -2x=4. Divide both sides by negative 2 to get X=-2. So the solution would be (-2,1)
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Answer:

The Answer is D

Step-by-step explanation:

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loris [4]

(a) It looks like the ODE is

<em>y'</em> = 4<em>x</em> √(1 - <em>y </em>^2)

which is separable:

d<em>y</em>/d<em>x</em> = 4<em>x</em> √(1 - <em>y</em> ^2)   =>   d<em>y</em>/√(1 - <em>y</em> ^2) = 4<em>x</em> d<em>x</em>

Integrate both sides. On the left, substitute <em>y</em> = sin(<em>t </em>) and d<em>y</em> = cos(<em>t</em> ) d<em>t</em> :

∫ d<em>y</em>/√(1 - <em>y</em> ^2) = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(1 - sin^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / √(cos^2(<em>t</em> )) d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

∫ cos(<em>t</em> ) / |cos(<em>t</em> )| d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

Since we want the substitutiong to be reversible, we implicitly assume that -<em>π</em>/2 ≤ <em>t</em> ≤ <em>π</em>/2, for which cos(<em>t</em> ) > 0, and in turn |cos(<em>t</em> )| = cos(<em>t</em> ). So the left side reduces completely and we get

∫ d<em>t</em> = ∫ 4<em>x</em> d<em>x</em>

<em>t</em> = 2<em>x</em> ^2 + <em>C</em>

arcsin(<em>y</em>) = 2<em>x</em> ^2 + <em>C</em>

<em>y</em> = sin(2<em>x</em> ^2 + <em>C </em>)

(b) There is no solution for the initial value <em>y</em> (0) = 4 because sin is bounded between -1 and 1.

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I THINK the answer is 40

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