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Alex Ar [27]
3 years ago
7

12

Mathematics
1 answer:
weeeeeb [17]3 years ago
8 0

Answer:

2x + y = -4

Step-by-step explanation:

standard form of equation of straight line is

ax+by = c

that  is terms containing x and y should be  on LHS and constant term should be  on RHS

______________________________________________

Given equation

y + 1 = - 2x - 3

lets bring -2x on LHS ,

add 2x on lHS and RHS

y + 1  + 2x = - 2x - 3 + 2x

=> y + 1  + 2x = -3

on lHS, 1 is there which constant term lets bring it on RHS

subtract 1 from both sides

y + 1  + 2x  - 1= -3 -1

y + 2x = -4

rearranging it

2x + y = -4  (Answer)

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For f(x)=4x+1 and g(x)=x^2-5 find (f-g)(x)
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Read 2 more answers
If 2(4x + 3)/(x - 3)(x + 7) = a/x - 3 + b/x + 7, find the values of a and b.
zmey [24]

Answer:

a=3 and b=5.

Step-by-step explanation:

So I believe the problem is this:

\frac{2(4x+3)}{x-3}(x+7)}=\frac{a}{x-3}+\frac{b}{x+7}

where we are asked to find values for a and b such that the equation holds for any x in the equation's domain.

So I'm actually going to get rid of any domain restrictions by multiplying both sides by (x-3)(x+7).

In other words this will clear the fractions.

\frac{2(4x+3)}{x-3}(x+7)}\cdot(x-3)(x+7)=\frac{a}{x-3}\cdot(x-3)(x+7)+\frac{b}{x+7}(x-3)(x+7)

2(4x+3)=a(x+7)+b(x-3)

As you can see there was some cancellation.

I'm going to plug in -7 for x because x+7 becomes 0 then.

2(4\cdot -7+3)=a(-7+7)+b(-7-3)

2(-28+3)=a(0)+b(-10)

2(-25)=0-10b

-50=-10b

Divide both sides by -10:

\frac{-50}{-10}=b

5=b

Now we have:

2(4x+3)=a(x+7)+b(x-3) with b=5

I notice that x-3 is 0 when x=3. So I'm going to replace x with 3.

2(4\cdot 3+3)=a(3+7)+b(3-3)

2(12+3)=a(10)+b(0)

2(15)=10a+0

30=10a

Divide both sides by 10:

\frac{30}{10}=a

3=a

So a=3 and b=5.

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2 years ago
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Answer:

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