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3 years ago

Solve this system of equations using the addition method.

Mathematics

1 answer:

damaskus [11]3 years ago

6 0

Eqt 1 = 4a-5b=7
eqt 2 = 4a+5b=17
eqt1+eqt2= 8a=24
a=3 then 4*3-5b=7
so b=1

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3 years ago

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Galina-37 [17]

Question:

Prove that:

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Answer:

Proved

Step-by-step explanation:

Given

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Required

Prove

$tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100)$

Subtract tan(10) from both sides

$- tan(10)+tan(10) + tan(70) + tan(100) = tan(10). tan(70). tan(100) - tan(10)$

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Factorize the right hand size

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Rewrite as:

$tan(70) + tan(100) = -tan(10)(1-tan(70). tan(100))$

Divide both sides by $1-tan(70). tan(100)$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = \frac{-tan(10)(1-tan(70). tan(100))}{1-tan(70). tan(100))}$

$\frac{tan(70) + tan(100)}{1-tan(70). tan(100)} = -tan(10)$

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$tan(A + B) = \frac{tan(A) + tan(B)}{1 - tan(A)tan(B)}$

So:

$\frac{tan(70) + tan(100)}{1 - tan(70)tan(100)}$ can be expressed as: $tan(70 + 100)$

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So:

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Because RHS = LHS

Then:

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