Simultaneous Equations

Probability

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Jan 26, 2012
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I have worked on this problem most of the day, I really wanted to solve it but am struggling to see what I am missing!

I have

2b = a + 7
3a = 2b - 1

I wanted to find "a" first but am unable.

Find b

a = 2b - 7

3(2b - 7) - 1 = 2b

6b - 21 - 1 = 2b

6b - 2b = 22

4b = 22

4b / 4 = 22 / 4

b = 5.5

Now I know that b = 5 and I now that a = 3, but unless I mix up the arithmetic and get the sign conventions deliberately wrong I just can't get anywhere near the right values?
 
I have worked on this problem most of the day, I really wanted to solve it but am struggling to see what I am missing!

I have

2b = a + 7
3a = 2b - 1

I wanted to find "a" first but am unable.

Find b

a = 2b - 7

3(2b - 7) - 1 = 2b

6b - 21 - 1 = 2b

6b - 2b = 22

4b = 22

4b / 4 = 22 / 4

b = 5.5

Now I know that b = 5 and I now that a = 3, but unless I mix up the arithmetic and get the sign conventions deliberately wrong I just can't get anywhere near the right values?

Your approach was fine. You just made a sign error when you "moved" the "1". You needed to add 1 to each side of the equation, so you should have had a +1 on the left side, not a -1.

Another approach:
Let's look at your two equations:

2b = a + 7
3a = 2b - 1

Notice that they both have a "2b" term in them. The first equation is written in terms of "2b", so that means we have an easy substitution to do. We know that 2b equal "a + 7" from the first equation, so we just substitute that into the second equation in place of 2b:

3a = 2b - 1
3a = a + 7 - 1

Now you have an equation that only has "a" in it.
 
Your approach was fine. You just made a sign error when you "moved" the "1". You needed to add 1 to each side of the equation, so you should have had a +1 on the left side, not a -1.

Another approach:
Let's look at your two equations:

2b = a + 7
3a = 2b - 1

Notice that they both have a "2b" term in them. The first equation is written in terms of "2b", so that means we have an easy substitution to do. We know that 2b equal "a + 7" from the first equation, so we just substitute that into the second equation in place of 2b:

3a = 2b - 1
3a = a + 7 - 1

Now you have an equation that only has "a" in it.

Thanks I can see where I was going wrong now
 
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