Maths questions. Hit me please.

Bikerman2019 · 09-04-2020 2:04pm #1

So, I am really trying to pull up my algebra for my electronics course and hoping a few folks here can help me understand stuff. I assume it is allowed after reading the rules.

So here is the first.

1/2 (y-6) = 7

I understand to get rid of 1/2, multiply inside the bracket and both sides by 2 to give y-12=14

Yet mathopolis says it gives y-6=14 but in the same breath this on google for a similar equation says different procedures

2(x+3)=10
2x+6=10
x=2

So why is the first equation not multiplying both terms inside the bracket, yet the second does?

Thanks LOTS more to come.

timsey tiger · 09-04-2020 2:22pm

Your issue is with bomdas. Multiplication and division have the same priority, so after this we read from left to right.
1/2×(y-6)=7
0.5×(y-6)=7 (now multiply both sides by 2)
y-6=14 (now add 6 to both sides)
y=20

or alternatively, use the distributive law to multiply inside the brackets by 0.5

0.5y-3=7
0.5y=10
y=20

Bikerman2019 · 09-04-2020 2:28pm

timsey tiger wrote: »

Your issue is with bomdas. Multiplication and division have the same priority, so after this we read from left to right.
1/2×(y-6)=7
0.5×(y-6)=7 (now multiply both sides by 2)
y-6=14 (now add 6 to both sides)
y=20

or alternatively, use the distributive law to multiply inside the brackets by 0.5

0.5y-3=7
0.5y=10
y=20

When you multiply something outside a bracket, by a term inside a bracket, do you not do EVERYTHING inside the bracket?

ChippingSodbury · 09-04-2020 2:41pm

If you are multiplying or divide something in a equation, the same operation has to be applied to both sides of the =

e.g.
3(4 + 5) = 9 x 3

is the same as dividing BOTH sides by 3
(4 + 5) = 9

is the same as (each term on both sides is divided by three)
3 x 4 + 3 x 5 = 9 x 3 --> 3 x 4 / 3 + 3 x 5/ 3 = 9 x 3 / 3

Muir · 09-04-2020 2:52pm

So

1/2 (y-6) = 7

Is the same as

y/2 - 6/2 = 7

And when you multiply that by 2 you get

y - 6 = 14

The half outside is going to be multiplied by everything inside the bracket, so we only need to deal with the half outside to have the same result. Having the half outside is factorising, so you can multiply it in first to understand why it makes sense. But once it makes sense, it's quicker to leave the half outside and just get rid of it in one go, particularly if you move on to longer & more complex equations later.

I hope that makes sense!

timsey tiger · 09-04-2020 2:57pm

Bikerman2019 wrote: »

When you multiply something outside a bracket, by a term inside a bracket, do you not do EVERYTHING inside the bracket?

Yes

A*(B+C)=AB+AC

This is known as the distributive law.

Just to be clearer, your issue is with the "punctuation" of maths. The two scenarios you mention are the same, you are mixing up the order in which things happen.

You ask why you don't multiply inside the brackets by two? We don't do this here, because the order of the calculation based on the way you have written it is from left to right. First you must divide the 1 by the 2 and then multiply this by the brackets.

Adding brackets to make the order clearer

1/2(y-6) means the same thing as (1/2)×(y-6).

It is not the same as 1/(2×(y-6)).

TheChizler · 09-04-2020 3:01pm

But here you don't even have to look inside the brackets for this bit, the 2 and the 1/2 cancel (or combine to form 1 more accurately).

So
2*1/2 (y-6) = 2*7

1*(y-6)=14

timsey tiger · 09-04-2020 3:29pm

TheChizler wrote: »

But here you don't even have to look inside the brackets for this bit, the 2 and the 1/2 cancel (or combine to form 1 more accurately).

So
2*1/2 (y-6) = 2*7

1*(y-6)=14

True, however this does not really help the op in understanding why he can't do what he is proposing or why it is different from the second example he has given.

He wants to learn how to fish, not just have a fish dinner.

09-04-2020 4:02pm

Bikerman2019 wrote: »

I understand to get rid of 1/2, multiply inside the bracket and both sides by 2 to give y-12=14

This is the part that's wrong. You can get rid of the half by either multiplying the half by both terms inside the bracket or multiplying both sides by 2. You don't do both

Bikerman2019 wrote: »

2(x+3)=10
2x+6=10
x=2

So why is the first equation not multiplying both terms inside the bracket, yet the second does?

Because they only multiplied inside the brackets. They did nothing to the right-hand side

For equations like this, you can do anything you like, provided you do the exact same thing to both sides. If you're operating on a single side, you can only work with the exact information you have.

Say you had:
10 - 7 + 25 = 15 - 3 + 16 [Trivially true]

You can apply any amount of operations to both sides equally:
(1093125 * 457 / 21) * (10 - 7) = (1093125 * 457 / 21) * (15 - 12) [Still true, and you don't need to work it out to know that]

But if you want to work out one side at a time, you can't apply anything that isn't already there.

Maths questions. Hit me please.

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