4. System of equations
A system of linear equations are 2 or more (straight
line) equations that share variables. Since we typically
deal with 2 variables in algebra, the systems of
equations typically share x and y
When we solve a system of equations, we look to see if
the equations share a common (x, y) pair where they
intersect. That is the solution and if there is no
intersection, there is no solution
5. Here is a graph of a system of
equations that intersects:
y = 3x + 1
y = x – 2
What if we don’t have a graph and
have to find solutions from just 2
equations? One way we can solve
this is through elimination
7. To solve a system of equations using elimination, we first make sure that both
equations we are dealing with are in standard form
Here is the system
5x + 8y = 13
7x - 8y = 2
If there are any fractions in your equations, make sure to manipulate them so that
the fractions aren’t there anymore and that the coefficients are whole numbers!
The goal of this method of solving is to effectively “eliminate” one of the variables
so we can solve for the other one
Working through an example
8. The first step would be to look at the x values and the y values to see if one pair has
the same coefficient, but one value is positive and the other is negative. We see that
this is true for the y values because one of them is +8y and the other is -8y. Now we
add the x values, the y values, and the constants straight down
5x + 8y = 13
7x - 8y = 2
12x = 15
+8y and -8y become 0
9. So we now have 12x = 15. Solve for x and you get x = 5/4 (simplified). Plug x into either
equation to get the y value and we get:
5(5/4) + 8y = 13
25/4 + 8y = 52/4
8y = 27/4
y=27/32
11. If we have a system of equations like
3x + 5y = 15
8x + y = 2
We would have to manipulate the equations so that we would
get a pair with the same coefficient but opposite sides
What do we do if we DON’T have a pair with the same
coefficient but opposite signs?
12. Let’s do it for y. to get -5y as our y term for the second equation,
we would need to multiply not just y, but the whole second
equation by -5
-5(8x + y = 2) becomes -40x -5y = -10
Now if we have
3x + 5y = 15
-40x – 5y = -10
You can now eliminate the y term and solve for x first, y next,
and then you’re done
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