How many solutions does the following system of equations have 2y=5x+4 and y=3x+2

Hello Jen

2y = 5x + 4
y = 3x + 2

We gonna solve y= 3x + 2 for y
First, let start by substitute 3x + 2 for y in 2y = 5x + 4
2y = 5x + 4
2(3x + 2) = 5x + 4
6x +4 = 5x +4
Common terms
6x - 5x = 4 - 4
x = 0
We have x so we gonna use the value of x to find y
Substitute 0 for x in y =3x + 2
y = 3x + 2
y= (3)(0)+2
y = 0 + 2
y = 2

Answers : x = 0 and y = 2

There are two solutions in this system of equation.

I hope that helps!

The question simply requires some visualisation, you must think about what the graphs of the two equations will look like and work out how many times the two graphs will intersect. Sketching may be useful.
You should recognise that both eq'ns in this question are linear (straight line graphs) so they can only either intersect once or not at all. THIS IS ONLY THE CASE IF BOTH EQUATIONS BEING DEALT WITH ARE LINEAR (i.e. no powers of x or y or any other variable that are above 1).
What you should do is ignore all the numbers that are not coefficients of letters (i.e. for ax², a is the coefficient) and rearrange to get in terms of y.
In this case:
y = 5/2x = 2.5x and y = 3x,
You can easily tell that the coefficients of x are not are not the same, this means the graphs will definitely intersect.
If the coefficients of x are the same, this means the line graphs of the two eq'ns are parallel and so will never intersect.
The x-values of every point of intersection are regarded as solutions.

We need to solve simultaneously:
There are two methods that come to mind immediately, I will give my preferred one.
I would rearrange both to give in terms of y, the second eq'n is already in terms of y so we just have to deal with the first one.
2y = 5x + 4 --> y = (5x + 4) / 2

Now, simply equate and solve the two equations, eliminating y:
(5x + 4)/2 = 3x + 2          (×2)
5x + 4 = 6x + 4               (Rearrange)
x = 0
the y-coordinate of the point of intersection is found by substituting the found x-value into either but only one of the original equations:
y = 3(0) + 2 = 2
Only Point of Intersection: (0, 2)