y=- 13x+7 & (I) y=- 13x-2 & (II)
In this system, we can see that the y-variable in both equations is already isolated. This means that the substitution method might be the easiest method to solving the system. When using this method, we follow three steps.
Isolate a variable in one of the equations.
Substitute the expression for that variable into the other equation and solve.
Substitute this solution into one of the equations and solve for the value of the other variable.
Both expressions equal to y are rather simple, so it does not matter which we substitute into the other. Let's choose Equation (I) for our initial substitution.
Solving this system of equations resulted in a contradiction — -2 can never be equal to 7. Therefore, the lines are parallel and do not have a point of intersection.
b Let's take a look at the next system.
y=2x+3 & (I) y=x^2-2x+3 & (II)
Like in Part A, at least one of the variables has a coefficient of 1 in each equation in the system. Let's use the substitution method again. Since the expression equal to y in Equation (I) is simpler, we will use that for our initial substitution.
We have two possible values of x. Now we need to substitute x=0 and x=4 into either one of the equations in the given system to find the corresponding values of y. Let's use the first equation from the original system and substitute x=0 into it.
One of the solutions, or points of intersection, to this system of equations is the point (0,3). To find the other one, let's substitute x= 4 into the first equation and simplify.
