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andriy [413]
3 years ago
13

Please help me out :((

Mathematics
1 answer:
olga_2 [115]3 years ago
6 0

Answer:

1) 5*x + 6 = 2 + 3*x

5*x - 3*x = 2 - 6

(5 - 3)*x = -4

2*x = - 4

x = -4/2 = -2

x = -2

We have only one solution.

2) 2*(6 - 2*y) = -1*(4*y - 9)

12 - 4*y = -4*y + 9

12 - 9 = -4*y + 4*y = 0

3 = 0

This is absurd, so this equation has no solution.

3) 2*z - 6 = 2*(z + 2) - 10

2*z - 6 = 2*z + 4 - 10

2*z - 6 = 2*z - 6

Here we have the exact same thing on both sides of the equation, then we have infinite solutions, this happens because z can take any value, and the equation will be true always.

4) We want to have no solutions, so we need to end with something like:

1 = 6

so, we start with:

5*x + 1 = 5*x + A

We want to find the value of A.

First, we can subtract 5*x in both sides, so we get:

1 = A

Now we just need to take A different than 1, for example, if A  = 2, then:

1 = 2

In this case the equation has no solutions, then we have:

5*x + 1 = 5*x + 2

5) We want to have one solution:

3*x - 3 = A*x + 11

We want to end with something like x = k

Let's solve the equation for x:

3*x - 3 = A*x + 11

3*x - A*x = 11 + 3

(3 - A)*x = 14

if we take A = 2, then:

(3 - 2)*x = 14

x = 14

So this equation has only one solution, the equation is:

3*x - 3 = 2*x + 11

6) We want to have infinitely many solutions, then we need to have the same thing on both sides, then we need to have an equation like:

8*x - 7 = 8*x - 7

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