Two equations are shown: Equation 1: 3 4 (x−12)=12 Equation 2: 3 4 y−12=12 Solve each equation. Then, enter a number in each box to make this statement true. please help

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Two equations are shown: Equation 1: 3 4 (x−12)=12 Equation 2: 3 4 y−12=12 Solve each equation. Then, enter a number in each box to make this statement true. please help

Mathematics
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please help
@Vocaloid please help
Do you mean \(\dfrac{3}{4}(x - 12) = 12\) ?

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yes
Then the first thing you need to do is to get rid of the 3/4 from the left side. To do that, multiply both sides by 4/3
do i multiply 4/3
\(\dfrac{4}{3} \times \dfrac{3}{4}(x - 12) = \dfrac{4}{3} \times 12\)
\(\dfrac{\cancel{4}}{\cancel{3}} \times \dfrac{\cancel{3}}{\cancel{4}}(x - 12) = 16\)
Now you have simply: \(x - 12 = 16\)
what is the value of X
Now what do you need to do to get rid of the -12 on the left side?
Since 12 is being subtracted from x, you do the opposite operation, which is add 12. You must do the same operation to both sides of an equation, so you add 12 to both sides.
\(x - 12+12 = 16+12\) \(x = 28\)
that mean the value of X =28
Yes.
For the second equation, you have y instead of x, but also no parentheses, right?
then what is the value of Y
Is this the second equation? \(\dfrac{3}{4}y - 12 = 12\)
y = 12
Are you sure?
no
i think it is cuz u have to divide
Let's do it. Did I write the second equation correctly? Is the second equation like the first equation except you have y instead of x and no parentheses?
yes
\(\dfrac{3}{4}y - 12 = 12\) First, we need to get rid of the -12, so we add 12 to both sides. \(\dfrac{3}{4}y - 12 + 12 = 12 + 12\) \(\dfrac{3}{4}y = 24\) Do you follow so far?
yes
srry took a 5 min nap
Now we need to get rid of the 3/4 multiplying x, so we divide both sides by 3/4. Dividing by a fraction is the same as multiplying by its reciprocal. We now multiply both sides by 4/3, the reciprocal of 3/4.
multiply 4/3
\(\dfrac{\cancel{4}}{\cancel{3}} \times \dfrac{\cancel{3}}{\cancel{4}}y = \dfrac{4}{3} \times 24\) \(y = 32\)
As you can see, having or not having the parentheses in the original equation makes a difference in the answers.
thanks alot
You're welcome.
i have one more
can you help?

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