sytem of equations by method of elimination

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sytem of equations by method of elimination

Postby Galaxy » Thu 14 April, 2011 12:40

solve system of equation by elimination
3r-5s=-37
5r+3s=29

ok lets name them
3r-5s=-37 EQ1
5r+3s=29 EQ2

we want to make something in EQ1 to be the same as EQ2 so I think 5s in EQ1 and 3S in EQ2 are easy to make the same. So multiply EQ1 by 3 and EQ2 by 5 and we have:

in EQ1
3(3)r-(3)5s=-37(3)
9r-15s=-111

in EQ2
5(5)r+3(5)s=29(5)
25r+15s=145

now because both middle terms are the same but different sing (+ and -) we can add them to get rid of them:
9r-15s=-111
25r+15s=145
--------------------
34r=34

divide both side by 34 and we got `(34r)/34=34/34` which becomes `r=34/34=1`
so we found r=1

now find to find s we use r=1 in any of our equation EQ1 or EQ2 and here I'll use EQ2;
5r+3s=29 sub in the r=1 5(1)+3s=29 which is 5+3s=39. Now move 5 to other side by subtracting 5 from both side:

`5-5+3s=39-5` and we get: `3s=34` to have only s on the left side, divide both side by 3 we have `(3s)/3=34/3` and 3s on the left cancel out and we are left with:
`s=24/3=8`

so r=1 and s=8

now if we test r=1 and s=8 in both equation it should give the same answer:
in EQ1
3r-5s=-37
3(1)-5(8)=-37
3-40=-37
-37=-37 ✓

in EQ2
5r+3s=29
5(1)+3(8)=29
5+24=29
29=29 ✓

and indeed r=1 and s=8
Galaxy
 
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Joined: Sat 19 March, 2011 14:54

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