# question on SIMULTANEOUS EQUATIONZ. (1 Viewer)

##### New Member

Studying simultaenous equations and I still cant find the naswer to my question.

How do you know which variable x or y in either equation to isolate/ use for the subsitution.

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#### Kangaaroo

##### Member
You may isolate either variable, it doesn't matter. In this case, though, don't you think elimination would be better?

We can multiply the first equation through by 2, and get a new set of equations:

2x + 10y = -26 -- 1
2x - y = 7 -- 2

Subtracting 2 from 1, we have:
10y -(-y) = -26 - 7
11y = -33
y = -3

Now we have y, we can sub into equation 2 (or 1, if you prefer):
2x -(-3) = 7
2x + 3 = 7
2x = 4
x = 2

x = 2
y = -3

##### New Member
You may isolate either variable, it doesn't matter. In this case, though, don't you think elimination would be better?

Lets say I isolate y in equation 1. The answer is wrong. So there must be a way of selecting the right one.
Same for Isolating 2x in equation 2.

Also, I prefer substitution.

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#### fan96

##### 617 pages
A system of simultaneous equations is just a set of conditions. When we solve a system like this, we're simply finding and using different equations that represent the same conditions.

Point being, whether you choose to do this step or that step first doesn't matter, because nothing you're doing actually changes the conditions or solutions of the system - it's just to make it easier for you to solve.

If you're finding inconsistencies like you've described, then either your system has no solution or you've made an arithmetic error somewhere.

(The system does have a solution)

Isolating $\bg_white y$ in eq. $\bg_white (1)$,

$\bg_white (1) \to 5y = - 13-x$

$\bg_white (2) \to 10x - 5y = 35$ (multiplying the original by 5)

Substituting $\bg_white (1)$ into $\bg_white (2)$,

$\bg_white 10x+13+x=35$

$\bg_white \iff x = 2$

##### New Member
A system of simultaneous equations is just a set of conditions. When we solve a system like this, we're simply finding and using different equations that represent the same conditions.

Point being, whether you choose to do this step or that step first doesn't matter, because nothing you're doing actually changes the conditions or solutions of the system - it's just to make it easier for you to solve.

If you're finding inconsistencies like you've described, then either your system has no solution or you've made an arithmetic error somewhere.

(The system does have a solution)

Isolating $\bg_white y$ in eq. $\bg_white (1)$,

$\bg_white (1) \to 5y = - 13-x$

$\bg_white (2) \to 10x - 5y = 35$ (multiplying the original by 5)

Substituting $\bg_white (1)$ into $\bg_white (2)$,

$\bg_white 10x+13+x=35$

$\bg_white \iff x = 2$
so what decision do you make when selecting what to isolate for substitution. Is it random?

#### Kangaaroo

##### Member
so what decision do you make when selecting what to isolate for substitution. Is it random?
Well it can be either, but generally you'd simply choose the easiest one.

#### fan96

##### 617 pages
so what decision do you make when selecting what to isolate for substitution. Is it random?
If one of them is clearly more convenient, then pick that one.

For example, if we have the system

$\bg_white \begin{cases} 2x+7y &= 4 \\ 4x + 15y &= 3\end{cases}$

Then it would be smarter to eliminate $\bg_white x$ by doubling the first equation.

The alternative is to multiply equation 1 by $\bg_white 15$ and equation 2 by $\bg_white 7$ ... a much harder task.

But don't worry about it too much - at the end of the day both methods get you to the same conclusion.

#### JupiterPie

##### New Member
x + 5y = -13 --- (1)
2x - y = 7 --- (2)

From eq. (2)
y = 2x - 7 --- (3)

Putting this value of y in eq. (1)
x + 5(2x -7) = -13
x +10x - 35 = -23
11x = -13+35
11x = 22
x = 22/11
x = 2 ----- (A)

Putting x = 2 in (3)
y = 2(2) - 7
y = 4 - 7
y = -3 -----(B)

#### Roy G Biv

##### Member
so what decision do you make when selecting what to isolate for substitution. Is it random?
Doesn't matter at all Just choose whichever one you want. Once you gain experience, you'll start to gain a preference as to which you find easier, but really, it makes no difference.