# Thread: Having Trouble Solving An Algebra Equation

1. ## Having Trouble Solving An Algebra Equation

question is:

SOLVE:

4x = 5 = 3(2x+7)

i havent seen one like this yet, what with all the equal signs

2. Originally Posted by RyanCouture
question is:

SOLVE:

4x = 5 = 3(2x+7)

i havent seen one like this yet, what with all the equal signs

where did this equation come from? it makes no sense. there is no real x such that 4x = 5 and 3(2x + 7) = 5

i'm thinking it should be two equations.

4x = 5
=> x = 5/4

and

3(2x + 7) = 5
=> x = -8/3

3. yeah it doesnt make any sense

theres no way to re arrange it or anything?

4. Originally Posted by RyanCouture
yeah it doesnt make any sense

theres no way to re arrange it or anything?
like i said, the only thing that seems to make sense to me is to have two separate equations. where did this equation come from anyway? was it one that you made up, or was it given to you. tell me the question that you started with

5. it was given to me in a math book:

2. Solve the following equations:

a) 4x=5=3(2x+7)

b) 7x^3+23x+6=0

c) -64z^2 = -121

thats the whole segment

6. Originally Posted by RyanCouture
it was given to me in a math book:

2. Solve the following equations:

a) 4x=5=3(2x+7)

b) 7x^3+23x+6=0

c) =64z^2 = -121

thats the whole segment
i think you made a typo. what hinted that to me is the last equation you typed. you had =64z^2 = -121. it makes no sense to have the = in front of the 64 like that. are you sure it shouldn't be a minus sign, as in a - ?

look at the book again, maybe you should have: 4x - 5 = 3(2x + 7) and -64z^2 = -121

7. lol

but the first one we were talking about is typed correctly, this math course that im doing (its a mail in one) has had MANY typos so far, so i would be surprised if this is one.

8. Originally Posted by RyanCouture
lol

but the first one we were talking about is typed correctly, this math course that im doing (its a mail in one) has had MANY typos so far, so i would be surprised if this is one.
yea, its probably a typo. treat the first = sign as a minus. can you do that new equation?

9. i think so

do i just expand, then collect like terms

4x-5=3(2x+7)

4x-5=6x+21

4x-6x = 5+21

-2x = 26

x = 26/-2

x = -13

??

what do you think?

10. Jhevon ?

how do i go about c)

-64z = -121

thanks

11. you are right for #1.

For $-64z = -121$ divide both sides by $-64$.

So: $-\frac{64z}{-64} = -\frac{121}{-64}$

$z = -\frac{121}{-64} \approx 1.89$

Remember that a negative number divided by a negative number is a positive number. A negative number multiplied by a negative number is a positive number.

12. whoops, i made a mistake

the question for c) was:

-64z² = -121

forgot the square

do i go about it the same way??

13. If the equation is $-64z^{2} = -121$, then:

$-\frac{64z^{2}}{-64} = -\frac{121}{-64}$

$z^{2} \approx 1.89$

$z \approx \pm \sqrt{1.89} \approx \pm 1.37$

14. Originally Posted by tukeywilliams
If the equation is $-64z^{2} = -121$, then:

$-\frac{64z^{2}}{-64} = -\frac{121}{-64}$

$z^{2} \approx 1.89$

$z \approx \pm \sqrt{1.89} \approx \pm 1.37$
what does that + square root mean?

the answer is whatever the square root of 1.89 is?

15. ohh i see, nevermind