Some algebra 1 questions 10pts!

A)Solve: -12<3x-5<-4
Answer will be in {x_
B)Determine wether (-8,-3) is a soution of 3x-8y=-2

C)Solve using the multiplacation principle: -1/2x=-3/4

D)find the domain of the function: g(x)=7/10-9x

Thank you so much!

A) -12 < 3x - 5 < -4.

We want to get the x in the middle all by itself, so we need to get rid of the -5 and the 3. We do this by adding 5 to each side of the equation (the left, the middle, and the right), and dividing each side by 3:

Adding 5: -7 < 3x < 1 --> Dividing by 3: -7/3 < x < 1/3. Done!

B) In the brackets (-8, -3), the -8 refers to the x value, and the -3 refers to the y value. So we take these numbers and substitute them for the x and y variables in the equation:

3x - 8y = -2 --> 3(-8) - 8(-3) = -2 --> -24 + 24 = -2 --> Clearly, (-8, -3) is NOT a solution, because -24 + 24 = 0, not -2.

C) If we multiply both sides of the equation by -2, this will "cancel out" the -1/2 coefficient on the x, leaving us with: x = -3/2

D) Domain means all the possible x values we could use in the equation that will give us a positive solution, a negative solution, or 0. If you think about it, ANY "real" value of x will fit these criteria, so the domain of the function is all the real values of x, or: X = R

A. -12<3x-5<-4
-7<3x<1
-7/3
B. 3x-8y=-2
3(-8)-8(-3)=-2
-24+24=2
0=2
Statement false, therefore coordinate is not solution of equation.

C -1/2x=-3/4
-1/2 (2/-1) x= -3/4 (2/-1)
x= -3/4 (2/-1)
x=-6/-4
x=6/4
x=1.5

D.Domain is between negative infinity and infinity. Domain could be 0,1,2,-1,-2

A) -12<3x-5<-4
First divide by 3

-4 Then add 5 to both sides
1
B) Plug in -8 for x and -3 for y
3(-8) -8(-3) = -2
-24+24=-2
0 Does not equal -2 so therefore that is not a solution of the equation 3x-8y=-2

C) multiply (-1/2x)=3/4 by -2
x=-3/2

D) All reals

As long as there is nothing being divided by 0, the domain will be all reals.