Like what you saw?
Create FREE Account and:
 Watch all FREE content in 21 subjects(388 videos for 23 hours)
 FREE advice on how to get better grades at school from an expert
 Attend and watch FREE live webinar on useful topics
Solving a System of Linear Equations in Two Variables  Problem 2
Carl Horowitz
Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upperlevel math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
Solving a system of linear equations in two variables. When we’re solving a system of linear equations we have two tools at our disposal; we have substitution and we have elimination.
Substitution is we’re just taking a variable from one equation and plugging it into the other one. Looking at the problem behind me, see if you can figure out if substitution is a good option. What I see if every single variable has a coefficient. So every x and every y has a coefficient and there are no common factors amongst the other terms with those coefficients. What I mean by that is basically if I wanted to solve for this x, well I would have to divide by two but I’m going to introduce a fraction in there. If I want to solve for any variable, we’re going to get fractions.
Most people tend not to like fractions so I would recommend trying to avoid substitution in this case. If you like fractions feel free to go right ahead, I tend not to be such a fan so I’m going to try the other method which is elimination.
The principal behind elimination is basically to get the same coefficient on any of our variables so that way when we add or subtract our equations together those coefficients disappear and we’re left with a single variable. So in doing that what we can do is multiply one or both equations by a number in order to get the coefficients the same.
Looking at this we have the option of either trying to get rid of the xs or the ys. We don’t really have too many common factors, we have 2 and 5, the smallest thing they have in common is 10 or 4 and 10, the smallest thing they have in common is 20. So it doesn’t really matter which one we want to get rid of I’m just going to say let’s go for the xs. We could go for the ys as well.
So multiplying the xs, we need both coefficients to be 10. Let’s multiply the top equation by 5, multiply the bottom equation by 2. In general I try to get my coefficients equal and opposite, I find that students make a lot less mistakes when you’re adding as opposed to subtracting. Subtracting you have to distribute that negative through and things can go wrong. I tend to find it a little bit easier to get your coefficients equal and opposite but you could subtract as well and as long as you do your signs alright you’ll be fine.
Distributing this 5 through, we have 10x minus 20y is equal to 15. Distributing the 2 through, 10x plus 20y is equal to 14. So now I always include the sign that I’m doing, makes sure I keep everything straight. So I want to add so my 10 and 10 cancel, but what happens is the 10s cancel, the 20s cancel so what I’m really left with is zero on this side and then 29 on the other. Now what does that mean?
We have zero is equal to 29. That’s not a true statement so what that tells me is that there is no point on these two lines that actually intersect. So this is false which tells me that there are no solutions. In other words these two lines are parallel. They're parallel, two parallel lines are never going to cross. Whenever you get a false solution, whenever you get zero is equal to 29 or some sort of statement that doesn’t make sense, typically that means you have no solution, the lines aren’t going to intersect.
So taking a system of linear equations, trying to figure out which method we want to use, substitution or elimination, to then solving and interpreting our results. In this case we have a false statement which means we have no solution.
Please enter your name.
Are you sure you want to delete this comment?
Carl Horowitz
B.S. in Mathematics University of Michigan
He knows how to make difficult math concepts easy for everyone to understand. He speaks at a steady pace and his stepbystep explanations are easy to follow.
i love you you are the best, ive spent 3 hours trying to understand probability and this is making sense now finally”
BRIGHTSTORM IS A REVOLUTION !!!”
because of you i ve got a 100/100 in my test thanks”
Sample Problems (18)
Need help with a problem?
Watch expert teachers solve similar problems.

Solving a System of Linear Equations in Two Variables
Problem 1 10,182 viewsSolve the system:
4x − 2y = 7x + 2y = 3 
Solving a System of Linear Equations in Two Variables
Problem 2 6,362 viewsSolve the system:
2x − 4y = 35x + 10y = 7 
Solving a System of Linear Equations in Two Variables
Problem 3 5,151 viewsSolve the system:
6x + 21y = 12x + 7 y = 2 2 
Solving a System of Linear Equations in Two Variables
Problem 4 710 views 
Solving a System of Linear Equations in Two Variables
Problem 5 672 views 
Solving a System of Linear Equations in Two Variables
Problem 6 662 views 
Solving a System of Linear Equations in Two Variables
Problem 7 541 views 
Solving a System of Linear Equations in Two Variables
Problem 8 598 views 
Solving a System of Linear Equations in Two Variables
Problem 9 504 views 
Solving a System of Linear Equations in Two Variables
Problem 10 534 views 
Solving a System of Linear Equations in Two Variables
Problem 11 480 views 
Solving a System of Linear Equations in Two Variables
Problem 12 333 views 
Solving a System of Linear Equations in Two Variables
Problem 13 350 views 
Solving a System of Linear Equations in Two Variables
Problem 14 313 views 
Solving a System of Linear Equations in Two Variables
Problem 15 320 views 
Solving a System of Linear Equations in Two Variables
Problem 16 335 views 
Solving a System of Linear Equations in Two Variables
Problem 17 282 views 
Solving a System of Linear Equations in Two Variables
Problem 18 324 views
Comments (0)
Please Sign in or Sign up to add your comment.
·
Delete