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3 years ago

7

How on the earth do you do this ????

2 answers:

Dafna1 [17]3 years ago

8 0

You basically solve one equation to get the answer to the other, which then will allow you to solve the previous equation to get your answer

julsineya [31]3 years ago

6 0

Alright, if it tells you to solve by substitution, you need to solve for either one of the variables (either x or y) in one of the equations. Let's say you solved for x. Then with the value of one of the variables (it will be an expression with y), you substitute it into the other equation in which you will get a value without variables as y. Now substitute that value y back into the first equation in which you solved for x to get the value of x.

So for 1)
5x + y = 0
y = -5x

Now plug -5x as y into the other equation -5x + y = -10
-5x -5x = -10
-10x = -10
x = 1

Now with x=1, plug that back into y=-5x.
y = -5(1)
y = -5

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3 years ago

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2 years ago

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3 years ago

What addition statement does this show?

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D. –5.2 + (–6.4) = –11.6

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2 years ago

(1+tanx)/(sinx+cosx)=secx

jek_recluse [69]

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ is proved

<h3><u>Solution:</u></h3>

Given that,

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$ ------- (1)

First we will simplify the LHS and then compare it with RHS

$\text { L. H.S }=\frac{1+\tan x}{\sin x+\cos x}$ ------ (2)

$\text {We know that } \tan x=\frac{\sin x}{\cos x}$

Substitute this in eqn (2)

$=\frac{1+\frac{\sin x}{\cos x}}{\sin x+\cos x}$

On simplification we get,

$=\frac{\frac{\sin x+\cos x}{\cos x}}{\sin x+\cos x}$

$=\frac{\sin x+\cos x}{\cos x} \times \frac{1}{\sin x+\cos x}$

Cancelling the common terms (sinx + cosx)

$=\frac{1}{c o s x}$

We know secant is inverse of cosine

$=\sec x=R . H . S$

Thus L.H.S = R.H.S

$\frac{1+\tan x}{\sin x+\cos x}=\sec x$

Hence proved

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3 years ago