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omeli [17]
2 years ago
14

A) 2x + 5y = 12,4x + 3y = -4substitution ​

Mathematics
1 answer:
Juliette [100K]2 years ago
7 0

let 2x+5y=12......... equation I

also,4x+3y=-4........ equation ii

now, multiply equation i by 2

we get,2(2x+5y=12)

or,4x+10y=24

now subtract i from ii

4x+10y=24

-4x+3y=-4

so,the answer be 7y=28

or,y=4

now putting the value of y in equation I we get,

4x+10y=24

or,4x+10.4=24

or, 4x=24-40

or, 4x=16

or,x=4

therefore, the value of x and y are 4and 4

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How to find variables of these #2 & #3
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So... hmmm if you check the first picture below, for 2)

we could use the proportions of those small, medium and large similar triangles  like  \bf \cfrac{small}{large}\qquad \cfrac{x}{12}=\cfrac{6}{x}\impliedby \textit{solve for "x"}
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\cfrac{small}{large}\qquad \cfrac{z}{18}=\cfrac{6}{z}\impliedby \textit{solve for "z"}
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now.. for 3) will be the second picture below


\bf \cfrac{large}{medium}\qquad \cfrac{x+10}{2\sqrt{30}}=\cfrac{2\sqrt{30}}{10}\impliedby \textit{solve for "x"}
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\textit{now, because you already know what "x" is, we can use it below}
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\cfrac{large}{small}\qquad \cfrac{z}{x}=\cfrac{x+10}{z}\impliedby \textit{solve for "z"}
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\textit{and let us use "x" again below}
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\cfrac{small}{medium}\qquad \cfrac{y}{10}=\cfrac{x}{y}\impliedby \textit{solve for "y"}

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So,

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