Question about linear equations with 3 variables?

Question about linear equations with 3 variables?

Postby Guest » Sat Oct 22, 2011 9:43 am

When asked to solve:
a + b = 3
-b + c = 3
a + 2c =10
......how do you solve for a variable if you cannot solve for the same variable (using elimination) for the 1st and 2nd and 2nd and 3rd equations? Or in other words, how would you solve this using elimination?
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Re: Question about linear equations with 3 variables?

Postby Math Tutor » Sun Oct 23, 2011 3:16 am

The first equation:
a + b = 3
so
b = 3 - a

The second equation:
-b + c = 3
c = 3 + b = 3 + (3 - a) = 6 - a


The third equation:
a + 2c =10
a + 2(6 - a) = 10
a + 12 - 2a = 10
-a = 10 - 12
-a = -2 /multiply the both sides by -1
a = 2

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Re: Question about linear equations with 3 variables?

Postby leesajohnson » Fri Jul 01, 2016 4:31 am

a+b = 3 ---- (i)
-b + c = 3 -----(ii)
a + 2c =10 ------(iii)

From the equation (i) we get
b = 3 - a
Now put the value of b into the 2nd equation
-(3-a) + c = 3
c = 3 + 3 - a
c = 6-a
Now we will put value of c into 3rd equation
a+ 2(6-a) = 10
a + 12 - 2a = 10
-a=10 -12
-a= -2
a=2

value of a = 2

put the value of "a" into equation 1st equation
2+ b =3
b= 3-2
b= 1

Now we will put the value of b into 2nd equation
-1 + c= 3
c= 3+1
c= 4

We have found out the value of a,b and c now and values are a = 2, b = 1, c = 4

leesajohnson
 

Re: Question about linear equations with 3 variables?

Postby Guest » Thu Nov 21, 2019 7:44 pm

How I would do this: add "a+ b= 3" to "-b+ c= 3" to get "a+ c= 6". The other equation is a+ 2c= 10. Subtracting a+ c= 6 from that gives c= 4. Then a+ c= a+ 4= 6 so a= 2 and a+ b= 2+ b= 3 so b= 1.
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