# Thread: Help Fast!! Solve equation

1. For the substitution I have no idea what to do because for the example you did for me it only had one variable on the top line at the right, then an equation on the right. The other one is two variables on the left and a number on the right.

For the elimination:

-7s=27
s=-27/7

and I get3.857142857, and I know that's not right. I was trying to follow your example.

=> - 5q = 15 ...................(2) - (1)
=> q = -15/5 = -3

But 4p + 2q = 10
so 4p + 2(-3) = 10
=> 4p - 6 = 10
=> 4p = 10 + 6 = 16
=> p = 16/4
=> p = 4

So p = 4, q = -3
And now I have to sleep. Hopefully I can get it in the morning.

2. Originally Posted by Rocher
Yeah the first question I did type out wrong. Sorry to make you go to the trouble of solving a wrong question. It was :

The least of three consecutive integers is divided by 10, the next is divided by 17, the greatest is divided by 26 and the sum of the quotients is 10.

Yeah, my teacher wants us to learn both elimination and substitution so that has to be substitution.
1. The least of three consecutive integers is divided by 10, the next is divided by 17, the greatest is divided by 26 and the sum of the quotients is 10.

Consecutive integers means they are one after the other. For example, 1,2,3 or 7,8,9 are three consecutive intergers.

So if the first integer is x, the second is x + 1 and the third is x + 2.

Of course, the least integer will be the first one, so we have:

x/10 + (x + 1)/17 + (x + 2)/26 = 10

It's kind of hard coming up with the LCD for 10, 17, and 26, so let's just make the LCD the multiple of all of them = 10*17*26 = 4420.

x/10 + (x + 1)/17 + (x + 2)/26 = 10

442x + 260(x + 1) + 170(x + 2)
------------------------------ = 10
.................4420

Now multiply both sides by 4420 to cancel the denomenator:

=> 4426x + 260(x + 1) + 170(x + 2) = 44200
=> 442x + 260x + 260 + 170x + 340 = 44200 ..........expanded the brackets
=> 872x + 600 = 44200
=> 872x = 44200 - 600 = 43600
=> x = 43600/872
=> x = 50

So the three integers are 50, 51, and 52

3. Originally Posted by Rocher
Can you help me solve two more? I thought I would get it if you showed me the steps for two of them but it turns out I was wrong >_<. I tried the elimination one first. I did this question :

3r-5s= -9
6r+2s= 18

So I followed what you did.

-7s= 27
s= -27(divided by)7. But then I got the answer as: 23.14285714.

Can you help me do that question as well as the one where you have to substitute:

2d-3c= 5
5d+2c = 22
3r-5s= -9 .....................(1)
6r+2s= 18 .....................(2)
I am going to eliminate r, so i need 6r in the top equation, i will multiply through by 2.

6r - 10s = -18 ...............(3) = (1) * 2
6r + 2s = 18..................(2)

12s = 36 .................(2) - (3)
=> s = 36/12 = 3

But 3r - 5s = -9
=> 3r - 5(3) = -9
3r - 15 = -9
3r = 15 - 9 = 6
=> r = 6/3 = 2

So r = 2, s = 3

I'm tired too, we'll se where we are tomorrow. And please by careful in the future with the typos.
Which reminds me, double check
2d-3c= 5
5d+2c = 22
for typos

4. Originally Posted by Jhevon
Which reminds me, double check
2d-3c= 5
5d+2c = 22
for typos
Nope, that's the question alright.

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