Linear Equations: Solving for 3 unknowns given 3 equations

In summary, when you plug in the values for a and b into the original equations, you get that they both equal 0.5. This means that the "b" terms in both equations are eliminated and the system of equations can be simplified to two equations with just two unknowns.
  • #1
kaydis
23
1
Homework Statement
I need to work out the values of a, b, and c
Relevant Equations
a+b=0
-2a-2c = -1
-a-3b+c= 1
a+b=0

so..
a= -b

-2a-2c = -1 = -2(-b)-2c = -1 = 2b-2c=-1

-a-3b+c= 1 = -(-b)-3b+c= 1 = b-3b+c=1 = -2b+c=1

i think this is right but i don't know where to go from here
 
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  • #2
kaydis said:
a= -b

-2a-2c = -1 = -2(-b)-2c = -1 = 2b-2c=-1

-a-3b+c= 1 = -(-b)-3b+c= 1 = b-3b+c=1 = -2b+c=1
Good work! You've been able to reduce the system of equations from 3 equations and 3 unknowns to 2 equations with 2 unknowns. What can you do to combine those last two equations to eliminate the "b" terms? :smile:
 
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  • #3
BTW, remember that after you have solved for a, b and c, it's a good idea to plug those back into the original 3 equations that you started with to be sure they are satisfied by the solutions you found. :smile:
 
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  • #4
berkeman said:
Good work! You've been able to reduce the system of equations from 3 equations and 3 unknowns to 2 equations with 2 unknowns. What can you do to combine those last two equations to eliminate the "b" terms? :smile:
could i add them?

2b + -2b = canceled out
and
-2c + c = -c
and
-1 + 1 = 0
so
-c=0
 
  • #5
Yep, good job. Now solve for b, then solve for a, and check that your values work in the original equations. Can you show us that? :smile:
 
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  • #6
I think i did it!
-c=0

-2a-2c=-1 = -2a-2(0)=-1 = -2a=-1 = -a=-1/2 or -0.5

SO a=0.5

-a-3b+0=1 = -a-3b=1 = -0.5-3b=1 = -3b=1.5 = -b=1/2 or 0.5

SO b=-0.5

when i plug them back in i get the following:

a+b=0 --------> 0.5+-0.5=0
-2a-2c=-1 --------> -2(0.5)-2(0)=-1
-a-3b+c=1 --------> -0.5-3(-0.5)+0=1
 
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  • #7
Looks good! :smile:
 
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  • #8
awesome, thank you so much for all the help! :smile:
 

Related to Linear Equations: Solving for 3 unknowns given 3 equations

1. What is a linear equation?

A linear equation is an algebraic equation that contains only terms with a degree of 1. It can be written in the form of y = mx + b, where m is the slope and b is the y-intercept.

2. How do you solve a linear equation with 3 unknowns and 3 equations?

In order to solve a system of linear equations with 3 unknowns and 3 equations, you can use the substitution method or the elimination method. Both methods involve isolating one variable in one equation and substituting it into the other equations until all 3 unknowns have been solved for.

3. Can there be more than one solution to a system of linear equations with 3 unknowns?

Yes, there can be more than one solution to a system of linear equations with 3 unknowns. This is known as an infinite number of solutions. This can happen when all three equations are essentially the same, resulting in multiple possible solutions.

4. Is it possible for a system of linear equations with 3 unknowns to have no solutions?

Yes, it is possible for a system of linear equations with 3 unknowns to have no solutions. This can happen when the equations are contradictory, meaning they have no common solution that satisfies all three equations. This is known as an inconsistent system.

5. Can a system of linear equations with 3 unknowns be solved graphically?

Yes, a system of linear equations with 3 unknowns can be solved graphically by graphing each equation on the same coordinate plane and finding the point of intersection. However, this method can be less precise and time-consuming compared to the substitution or elimination method.

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