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These arent the real problems..... these are examples I made up so I dont have to cheat....

Will you please help??!?!?! Please dont just give me the answer.. walk me threw it?!?!?!?!?

Then if so will you give me more so I can practice and remember how to do them?!?! (Thats optinal.. thx)

1) Y=5x 3x+y= -96

2) n=3p-4 2p- 3n= -2

3) m=4 (5-m)=50

4) 6-2c > 32 ( that one i dont even know how to do!!)

5) -6x < -36 (same)

6) 3x + 7 = 5x - 3

7) 7 = 5n - 12n

2006-09-10 06:46:15 · 10 answers · asked by ♥♥♥dannii 3 in Science & Mathematics Mathematics

10 answers

For questions 1 through 3, you must plug-in the number for the variable into the equation. The variable is the letter that stands for an unknown value, in the case of 1 through 3, the value for the variables are given for you. So, for example I will work out problem #1 for you.

Y = 5x - This means that the value for Y is 5x. Now, your equation is 3x + Y = -96. As I said above, you must plug in the value of the variable. Meaning, where the variable Y sits in the equation, you must substitute it with 5x. So, your new equation would be 3x + 5x = -96. Next, you must add like-terms, which simply means you must add 3x & 5x together, which equals 8x. Now that you've added the like terms, your new equation is 8x = -96. Now, all you've got to do is divide both sides by 8, so as to get the variable X alone. Therefore the answer to #1 is X = -12. For the remaining two questions, you are to follow the same steps.

For questions 4 & 5, an alternate approach is necessary. For example, I will work out #4 for you.

First, you must subtract 6 from both sides, as this is the first step that is necessary to get the variable C alone. In doing so, the equation will now be -2c > 26. Taking it in mind that you are trying to get the variable C alone in the equation, you must now divide by sides by -2. Thus, your answer is C > -13. For question 5, use the same approach.

6 & 7 are slightly similar to questions 1 - 3, only you are trying to solve from the variable X in #6 & variable N in #7. I will solve #6.

To begin, you must subtract the 7 from both the left side & right side of the equation, setting up the left side to hold only the variable X, making the equation 3x = 5x - 10. Now, so as to get the variable X on one side, you must subtract 5x from both sides. Making the equation -2x = -10. As stated above, you must now divide both sides by -2 so as to get the variable X by itself. Therefore, the answer the #6 is X = 5.

2006-09-10 07:18:24 · answer #1 · answered by Houston B 1 · 0 1

In each of the problems, you need to substitute variables so that you are can solve the equation using only one of the variables.

For example, on problem #1, since y=5x, try substituting the y in the second equation so that you get 3x + (5x) = -96. Then combine the terms and solve for x. It's pretty much the same process for the rest of the equations.

The only difference on #4 and #5 is that c and x can have more than one answer. Try pretending that the > is an equal sign for a second and solve the problem. The answer then is any number less than the one you found.

Hope this helps, good luck!

2006-09-10 13:59:17 · answer #2 · answered by cushdogjr 3 · 0 0

1. 3x+5x= -96
8x= -96
x= -16

2. can you write it more detailed is mixed

3. 5-4= 50
1=50 no solution

4. 6-2c greater than 32
-6 -6

-2c = 26
c is less than -13

5. -6x less than -36
x is greater than 6

6. 3x+7 = 5x-3
-3x -3x

7= 2x-3
+3 +3

10=2x
x=5

7. 7=5n - 12n
7=-7n
n= -1

2006-09-10 14:06:15 · answer #3 · answered by juliotelehit 2 · 0 0

1. (*) y = 5x, (**) 3x + y = -96
sub (*) into (**)
(**) 3x + y = -96
(**) 3x + 5x = -96
(**) 8x = -96
--> x = -12
sub x = -12 into (*)
(*) y = 5x
(*) y = 5(-12)
--> y = -60

For 2 and 3 you can use the same approach.

4. 6 - 2c > 32
(6 - 2c) - 6 > 32 - 6
-2c > 26
(-2c)/-2 < 26/-2 (Note: the inequality sign flips when you divide or multiply by -1)
c < -13

Use the same approach for 5

6. 3x + 7 = 5x - 3
3x + 7 - 5x = 5x - 3 - 5x
-2x + 7 = -3
-2x + 7 - 7 = -3 - 7
-2x = -10
x = 5

Use the same approach for 7.

Good luck!

2006-09-10 13:58:25 · answer #4 · answered by Jess 2 · 0 0

1) y=5x...equation (i)
3x+y= -96....equation(ii)

therefore......by substituting equation(i) in (ii) we get:->
3x+5x= -96
8x= -96
x= -12 and y=5*-12

2) n=3p-4...equation(i)
2p-3n= -2..''(ii)

substituting (i) in (ii) we get:->
2p-(3p-4)=-2
2p-3p+4=-2
-p+4=-2
-p=-6
p=6 and n=22

3) 20-4m-50=0
4m+30=0
m=30/4
m=7(1/2)

4) 6-2c-32 > 0
2c+32 < 0
2c < -32 (signs changed)
c < -16

5) 6x > 36 (inequality sign changed)
x>6

6) 3x+7-5x+3=0
10-2x=0
x=10/2
x=5

7) 7=5n-12n
7= -7n
-n=1
n= -1(signs changed)

2006-09-10 14:07:02 · answer #5 · answered by tonima 4 · 0 0

Let's tackle 4 and 5. These seem to be the hardest
for you.
6 -2c > 32.
Let's subtract 6 from both sides
-2c > 26
Now let's divide both sides by -2.
When you divide both sides of an inequality by a negativee
number you must reverse the sense of the inequality.
So, c < -13.

Number 5 follows the same rule. Divide both
sides by -6 and x > 6.
Hope that helps a bit.

2006-09-10 14:15:37 · answer #6 · answered by steiner1745 7 · 0 0

1. substitute the 5x from equation one into the second.3x+5x = -96....8x = -96 then devide by 8. same for #2.
#3 just seems wrong.
#4 subtract 6 from both sides....-2c> 26 then devide by -2.......careful when deviding by a neg.... it changes the inequality sign...... c < -13 same with #5.
#6 get your like terms together..... 7 = 5x-3x -3....7=2x-3......7-3= 2x ....5 = 2x ...x=5/2 same for # 7.

2006-09-10 15:44:29 · answer #7 · answered by who be boo? 5 · 0 0

1) Y=5x 3x+y= -96 ...... ok, so the y=5x is the key, you plug it into the equasion..... 3x+(5x)= -96...... 8x= -96.. then divide both sides by 8... x= -12

2- sameway.. plug it in...
3same

hope that helps... its been a while, but im pretty sure thats all correct.

2006-09-10 13:56:34 · answer #8 · answered by Jill 2 · 0 0

1.y=5x
3x+5x=-96
x=-12

2.2p-3n=-2
6p-9n=-6
6p-2n=8
7n=-14 n=-2 so p=-4

3.m=4
5-m=50 the sum is wrong

4.6-2c>32
-2c>26
-c>13
c<13

5.6x>36
x>6

6.3x+7=5x-3
-2x=-10
x=5

7.-7n=7
n=-1

2006-09-10 13:58:38 · answer #9 · answered by raj 7 · 0 0

=ask your math teacher

2006-09-10 13:48:38 · answer #10 · answered by Anonymous · 0 1

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