# Show that 5(3y-2) = 15y-10 for at least 5 values of y. I don't understand. Do I need to assign 5 different values to y? I am confused on what the question is asking specifically. Do I need...

Show that 5(3y-2) = 15y-10 for at least 5 values of y.

I don't understand. Do I need to assign 5 different values to y?

I am confused on what the question is asking specifically. Do I need to assign 5 different values to y. For example:

y = 2 and then do the entire problem from there. Or is there just one answer? I am confused when it comes to "for at least 5 values of y".

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### 2 Answers

Notice that if you open the brackets, the result will be the same with the expression to the right side, hence, if the expressions both sides are equal, then it is obvious that any value of y, substituted in equation, will give equal results.

`5(3y - 2) = 15y - 10 => 5*3y + 5*(-2) = 15y - 10`

Multiplying the coefficients and considering the rule of signs yields:

`15y - 10 = 15y - 10`

**There is no need to give values to y to evaluate the expressions both sides, because they are equal.**

**Sources:**

Yes we are going to assign at least five different values of y to prove the given equation.

So, let's assign values of y. Then, susbtitute it to the equation

`5(3y-2)=15y-10` .

When y=-1 When y=0,

`5(3(-1)-2)=15(-1)-10` `5(3*0-2)=15*0-10`

`5(-3-2)=-15-10` `5(0-2)=0-10`

`5(-5)=-25` `5(-2)=-10`

`-25=-25` (True) `-10=-10` (True)

When y=1 When y=4

`5(3*1-2)=15*1-10` `5(3*4-2)=15*4-10`

`5(3-2)=15-10` `5(12-2)=60-10`

`5(1)=5` `5(10)=50`

`5=5` (True) `50=50` (True)

When y=10

`5(3*10-2)=15*10-10`

`5(30-2)=150-10`

`5(28)=140`

`140=140` (True)

**Since all the five values of y result to a true condition, this proves that `5(3y-2)=15y-10` .**

**Sources:**