# solve by elimination

• Jan 18th 2009, 08:55 PM
annabananzaa
solve by elimination
5x-y=9
x-1.5y=-3

i dont even kno what it means by elimination. i know how to solve by substitution, but not elimination....
• Jan 18th 2009, 10:22 PM
Substitution
Hello annabananzaa
Quote:

Originally Posted by annabananzaa
5x-y=9
x-1.5y=-3

i dont even kno what it means by elimination. i know how to solve by substitution, but not elimination....

It's the same thing! When you substitute for x, you eliminate (= get rid of) it!

• Jan 18th 2009, 10:35 PM
annabananzaa
so i take the first problem, turn it into y=5x-9 and then just plug that into the first equation so it looks like x-1.5(5x-9)=9

would that be right?
• Jan 18th 2009, 10:43 PM
Substitution
Hello annabananzaa
Quote:

Originally Posted by annabananzaa
so i take the first problem, turn it into y=5x-9 and then just plug that into the first equation so it looks like x-1.5(5x-9)=9

would that be right?

Yes. Just be careful with the minus signs when you multiply out the brackets, won't you?

• Jan 18th 2009, 10:53 PM
annabananzaa
of course i will!

so this is what im coming up with step by step, and it doesnt seem right still....

x-1.5(5x-9)=9
x-7.5x+13.5=9
-6.5x+13.5=9
-6.5x=-4.5

-4.5/-6.5

x=.69?
• Jan 18th 2009, 11:15 PM
mr fantastic
Quote:

Hello annabananzaaIt's the same thing! When you substitute for x, you eliminate (= get rid of) it!

Crazily enough, it's not ..... Elimination and substitution are taught as two seperate methods.

One way of applying the elimination method:

\$\displaystyle 5x - y = 9\$ .... (1)

\$\displaystyle x - 1.5y = -3\$ .... (2)

Multiply equation (2) by -5:

\$\displaystyle -5x + 7.5y = 15\$ .... (3)

Now add equation (1) and equation (3) to eliminate \$\displaystyle x\$:

\$\displaystyle 6.5 y = 24 \Rightarrow y = \, ....\$

Now substitute the value of \$\displaystyle y\$ into any equation and solve for \$\displaystyle x\$.

This is what is taught in schools (so don't shoot me ....)
• Jan 18th 2009, 11:24 PM
annabananzaa
so i was right when saying my answer looked a little crazy?

im sorry, im still confused... i dont see where you got the 3 from in the begining and then i dont get why i would add the equations....
• Jan 18th 2009, 11:30 PM
mr fantastic
Quote:

Originally Posted by annabananzaa
so i was right when saying my answer looked a little crazy?

im sorry, im still confused... i dont see where you got the 3 from in the begining and then i dont get why i would add the equations....

What 3? I have labelled the equations (1), (2) and (3).

You add so that x gets eliminated.

I have (reasonably) assumed that you have some familiarity with the method ....

• Jan 18th 2009, 11:53 PM
annabananzaa
so would my first answer that i posted be correct or this new answer that i got after following the steps on the link??
x=3.69

iim tryingg really hard!
just not very good at math :(
• Jan 19th 2009, 12:25 AM
mr fantastic
Quote:

Originally Posted by annabananzaa
so would my first answer that i posted be correct or this new answer that i got after following the steps on the link??
x=3.69

iim tryingg really hard!
just not very good at math :(

If you do it how I showed you get y = 24/6.5 = 48/13 which is approximately equal to 3.69.

Now substitute and solve for x. You should get \$\displaystyle x \approx 2.538\$ (Note:an exact answer might be required).
• Jan 19th 2009, 12:45 AM
annabananzaa
ok. now i get it, i was forgetting the last step. thats what was really confusing me.
thank you so much for your patience!