How many Solutions are there for -6(3x+2)=2x-5-20x-7

Mathematics

1 answer:

jeka57 [31]3 years ago

5 0

First of all, expand the left hand side:

$-18x-12 = 2x-5-20x-7$

Sum like terms on the right hand side:

$-18x-12 = -18x-12$

As you can see, the equation turned to be an identity: it doesn't matter which value for x you are choosing: you're evaluating the same expression in both sides, so they will always be the same.

In other words, you're asking: which value for x should I choose so that -18x-12 and -18x-12 evaluate to the same number? Well, obvisouly any value for x will do.

So, there are infinitely many solutions to this equation, namely ever real number is a solution of this equation.

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A bank charges $0.25 per check and its monthly fees are $10. If Jerome issues 10 checks in a month, how much will a monthly chec

blsea [12.9K]

For this case, the first thing we must do is define variables.
We have then:
x: number of checks.
y: total cost.
We now write the linear function that represents the problem.
We have then:
$y = 0.25x + 10
$
Then, Jerome issues 10 checks in a month.
Therefore, we evaluate the function for x = 10.
We have then:
$y = 0.25 (10) + 10

y = 2.5 + 10

y = 12.5$
Answer:
a monthly checking account will cost him about:
y = 12.5 $

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Vlad1618 [11]

First or forth not sure

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3 years ago

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↔ ↔

MrRissso [65]

Answer:

1. $\dfrac{4}{3}$

2. $y=6x-11$

3. $4$

Step-by-step explanation:

The equation of the line that passes trough the points A(14,-1) and B(2,1) is

$\dfrac{x-14}{2-14}=\dfrac{y-(-1)}{1-(-1)},\\ \\\dfrac{x-14}{-12}=\dfrac{y+1}{2},\\ \\y+1=-\dfrac{1}{6}(x-14),\\ \\y=-\dfrac{1}{6}x+\dfrac{7}{3}-1,\\ \\y=-\dfrac{1}{6}x+\dfrac{4}{3}.$