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Aleks04 [339]
3 years ago
8

Can someone please help solve these simultaneous equations?

Mathematics
1 answer:
gtnhenbr [62]3 years ago
3 0

Answer:

a) x = 2, y = 1

b) x = 9, y = -9

Step-by-step explanation:

a)     2^{x+y} = 8

   ⇒ 2^{x+y} = 2^{3}

   ⇒ x+y = 3

and

      3^{2x-y} = 27

  ⇒ 3^{2x-y} = 3^{3\\

  ⇒ 2x-y = 3

By solving the above two equations, we get;

x = 2 and y = 1

b)     3^{y+x} = 9^{y+x}

   ⇒ 3^{y+x} = 3^{2y+2x}

   ⇒ x=-y

and

       2^{x-y} = 8^{x-3}

   ⇒ 2^{x-y} = 2^{3x-9}

   ⇒ 2x + y=9

By solving the above two equations, we get;

x = 9 and y = -9

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How is solving 2x+c=d similar to solving 2x+1=9 for x? How are they different? How can you use x+c=d to solve 2x+1=9?
valentinak56 [21]

Answer:

No

Step-by-step explanation:

We can't use 2x + c = d for solving 2x + 1 = 9 . Here we can find the value of x from 2x + 1 = 9.

2x + 1 =9\\=>2x=9-1\\=>2x=8\\=>x=4

Putting the value of x in 2x + c = d :-

2x+c=d\\=>2*4+c=d\\=>8+c=d

But still we don't know the exact value of c & d . If in the question at least we could have got one exact value of either c or d , then we could find out the value of other variable.

3 0
4 years ago
Side a=15 and side b=20 what is side c
Digiron [165]

Answer:

side c is 25

Step-by-step explanation:

6 0
3 years ago
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Please help would be very very appreciated
poizon [28]
115 ounces is the answer
8 0
4 years ago
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I have 2 thousands,12 tens,5 hundred and 9 one
12345 [234]

Answer:

2 120 500 9. so 2,629 would be the answer

7 0
3 years ago
In the system shown below, what are the coordinates of the solution that lies in quadrant I?
Arisa [49]

Answer:

(4,3)

Step-by-step explanation:

Given

x^2+y^2=25

x-y^2=-5

In order to solve the equations, from equation 2 we get

-y^2= -5-x

y^2=5+x

Putting the value of y^2 in equation 1

x^2+5+x=25

x^2+5-25+x=0

x^2+x-20=0

x^2+5x-4x-20= 0

x(x+5)-4(x+5)=0

(x+5)(x-4)=0

So

x+5=0  x-4=0

x=-5   x=4

Now for x=-5

x^2+y^2=25

(-5)^2+y^2=25

25+y^2=25

y^2=25-25

y^2=0

so Y=0  

And for x = 4

x^2+y^2=25

(4)^2+y^2=25

16+y^2=25

y^2=25-16

y^2=9

y= ±3

So the solution to the system of equations is

(-5,0) , (4,3), (4,-3)

The only solution that belongs to first quadrant is (4,3)

4 0
3 years ago
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