A system of two linear equations with the constants a, b, c, d, e, and f
is of the form ax plus by equals c, and second equation dx plus ey equals f.
What we are going to do is learn how to solve a
system of two linear equations by substitution and elimination.
It is also possible to solve these systems by graphing.
If we graph one linear equation and we graph a second one,
one of the possibilities in a rectangular coordinate plane
is that the two lines intersect at a single point with the coordinates (x, y).
This system is, because it has one solution, is called consistent.
Now that is one possibility. A second possibility is that the two lines may be parallel,
in which case there is no solution because the two lines do not intersect.
This is called an inconsistent system.
The third possibility is that the two lines are actually the same line.
In this case, we have an infinite number of solutions,
and this system is known as a dependent system.
This is the background material for a system of two linear equations.
We will look at our first example.
Solve by substitution or elimination.
We have the system 3x minus 6y equals 2, 5x plus 4y equals 1.
We have a choice as to which method we want to use.
Now because the coefficients of our x’s and y’s are greater than one,
substitution is not a good choice. I am going to use elimination,
and in elimination I need for the coefficients of the x’s to be able to cancel
or to give a zero when we add them.
In other words, I would need this to be 15x and this to be negative 15x,
or if I want to eliminate the y’s I would need
this term to be negative 12y and this one to be positive 12y.
The choice is yours as to which of the two variables you are going to eliminate.
Let’s write our system, and I think I will eliminate the y’s.
So, to get negative 12 I need to multiply by a positive 2,
and I will multiply both sides of the equation.
To get a positive 12 I am going to multiply both sides by positive 3.
Let’s be very careful as we distribute.
2 times 3x is 6x, 2 times negative 6y is negative 12y, and 2 times 2 is 4.
3 times 5x is 15x, 3 times 4y is 12y, 3 times 1 is 3.
Now we want to add. 6x and 15x is 21x, the y’s are eliminated, 4 and 3 is 7.
Solving now by dividing both sides by 21 we get 7 divided by 21,
but we reduce that to lowest terms, so we get x is a positive 1/3.
Now we need to find y, so we are going to take the value that we found for x,
substitute it into one of the equations. I am just going to choose the first equation.
3 times 1/3 minus 6y equals 2. 3 times 1/3 is 1.
Subtract the 1. Divide by the negative 6.
We get negative 1/6. So we see that our solution is the ordered pair (1/3, negative 1/6).
Now if we remember from our earlier discussion graphically what this means
in this system the lines that are represented here intersect at this one point,
so it is a consistent system. We will look at a second example.
Solve by substitution or elimination. 3x plus y equals 10. 12x plus 4y equals 14.
We have a y that has a coefficient of 1, so this system would be
very convenient for us to solve by substitution. So let’s try that method.
y equals, subtracting the 3x, y equals 10 minus 3x.
Now I am going to take that expression for y and substitute into the second equation.
So I will have 12x plus 4 times 10 minus 3x equals 14.
So what has happened is the y has been eliminated by substitution.
We will distribute the 4, 40 minus 12x equals 14.
We will combine the like terms. 12x minus 12x is 0. 40 equals 14.
In this example, not only was the y eliminated here, but now the x’s are eliminated.
We look at the remaining sentence, which we know is not possible,
so this system has no solution. Let’s circle that as our final answer.
Recall graphically, what does this look like?
Well this line is about here, with a y-intercept at 10 and a negative slope.
If we solve this equation for y, it has a negative slope and a y-intercept of 5/2.
Because the slopes are equal, we know that the lines are parallel.
On my graph they don’t look very parallel, but they are parallel.
We have an inconsistent system, which gives us no solution.
We will look at a third example.
Solve by substitution or elimination. We have the system
x plus 2y equals 4, 3x plus 6y equals 12.
Substitution, we could simply solve for x in the first equation
and then substitute into the second equation.
If we use elimination, we have several choices.
I think the choice that I will use is to multiply the first equation by negative 3.
Now of course you may recognize that
if you divided the second equation by 3 you would get x plus 2y equals 4.
So that gives us a hint that if we end up with the same two equations
what is going to happen with our solution.
Let’s just take that first equation and multiply the equation by negative 3.
So that is going to give us negative 3x minus 6y equals negative 12.
That becomes our first equation.
Second equation, 3x plus 6y equals positive 12.
When we add the x’s give us 0. The y’s, negative 6y plus 6y also gives us 0.
Then on the right side, we have negative 12 plus 12 which is 0.
This is what we suspected, that we just have a dependent system,
actually consistent and dependent, and our solution is going to be
an infinite number of ordered pairs.
Let’s go ahead and solve for y and see what that graph would look like.
2y would be negative x plus 4. y will be negative 1/2 x plus 2.
So quick graph. y- intercept 2, slope of negative 1/2.
We get that line, which would be the same for both.
We are going to write our solution as the set of ordered pairs (x, y)
such that, we have solved for y, y is negative 1/2 x plus 2
where x is any real number. Some students are always sometimes confused
because they think all real numbers will work,
but it is only the ordered pairs (x, y) where x is a real number and then y has this form.
So for example one solution is (0, 2), our y-intercept.
If we let x be 2, we would have y is negative 1/2 times 2, that would be,
oh let’s figure that out. That would be negative 1/2 times 2 plus 2.
That will give me negative 1 plus 2, or positive 1.
Those are just two of the solutions to this system of linear equations.
These three examples should show you how to solve any system
of two linear equations by substitution or elimination.