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3 years ago

2x+3y=13 4x-y=2 solve the simaltaneous equation

Mathematics

1 answer:

nekit [7.7K]3 years ago

5 0

Solve by substitution. We begin by manipulating the second equation (solving for y).

2x+3y=13

4x-y=-2 => -y=-4x-2 => y=4x+2

Now substitute "4x+2" for y into the first equation because y=4x+2.

2x+3y=13 becomes 2x+3(4x+2)=13

Now solve for x.

2x+3(4x+2)=13

2x+12x+6=13

14x=7

x=1/2

Now we can solve for y because we know what x is.

y=4x+2 => y=4(1/2)+2 = y=4

answer: x=1/2, y=4, intersection point (1/2,4)

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If sin A = 3/8, find the value of cosec A - sec A.

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Answer:

$\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

Step by step explanation:

$\text{Given that,}\\\\~~~~~~\sin A = \dfrac 38 \\\\\implies \sin^2 A = \dfrac 9{64}\\\\\implies 1 - \cos^2 A = \dfrac{9}{64}\\\\\implies \cos ^2 A = 1 - \dfrac 9{64}\\\\\implies \cos^2 A = \dfrac{55}{64}\\\\\implies \cos A =\pm\sqrt{\dfrac{55}{64}}\\ \\\implies \cos A = \pm\dfrac{\sqrt{55}}8\\\\$

$\implies \dfrac 1{\cos A} = \pm\dfrac{8}{\sqrt{55}}$

$\text{Now,}\\\\\csc A - \sec A\\\\=\dfrac{1}{\sin A}- \dfrac{1}{\cos A}\\\\=\dfrac 83 -\left(\pm \dfrac 8{\sqrt {55}} \right)\\ \\\text{Hence,}\\\\\csc A - \sec A = \dfrac 83 + \dfrac{8}{\sqrt{55}}\\\\\csc A - \sec A = \dfrac 83 - \dfrac{8}{\sqrt{55}}$

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2 years ago

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Veseljchak [2.6K]

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2 years ago

If 112 people attend a concert and tickets for adults cost $2.25 while tickets for children cost $and total receipts for the con

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Answer:

42 child and 70 adult

Step-by-step explanation:

Let a = number of adult tickets

c = number of child tickets

a+c = 112

2.25a + 1.75c = 231

Solve the first equation for a

a = 112-c

Substitute this into the second equation

2.25(112-c) + 1.75c = 231

Distribute

252 - 2.25c +1.75c = 231

Combine like terms

252 -.5c = 231

Subtract 252 from each side

252-252 -.5c = 231-252

-.5c = -21

Divide each side by -.5

-.5c/-.5 = -21/ -.5

c = 42

Now can find a

a = 112-c

a = 112-42

a = 70

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3 years ago

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