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zimovet [89]
3 years ago
14

Solving the following system using the elimination or the substitution method.

Mathematics
1 answer:
Lina20 [59]3 years ago
5 0

Answer and Step-by-step explanation:

For these system of equations, it would be easier to use the substitution method, as we are have one equation that is being equaled to a variable.

Plug the second equation into the value of b for the first equation.

5a + 3(-2a + 3) = 11

Distribute the 3.

5a - 6a + 3 = 11

Simplify.

-a + 3 = 11

Subtract 3 from both sides of the equation.

-a = 8

Divide both sides by -1.

a = -8

Now, we can plug this value into the second equation.

-2(-8) + 3 = b

Multiply -2 and -8 together.

16 + 3 = b

Add together 16 and 3.

19 = b

<u>a = -8</u>

<u>b = 19</u>

<u></u>

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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3 consecutive odd integers such that the sum of twice the first and three times the second is 55 more than twice the third
den301095 [7]

Answer:

Consecutive odd integers are 19 , 21 & 23

Step-by-step explanation:

Let the first 3 consecutive odd intergers be x , (x + 2) and (x + 4).

According to the question,

2x + 3(x + 2) = 2(x + 4) + 55

=  > 2x + 3x + 6 = 2x + 8 + 55

Eliminating 2x from both the sides,

=  > 3x + 6 = 63

=  > 3x = 63 - 6 = 57

=  > x =  \frac{57}{3}  = 19

So, the consecutive odd integers are = 19 , 21 & 23.

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Step-by-step explanation:

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