10x-5=3y
2x-3y=1
still lost help please
ok here are your tweo equations with the first one fixed
3 y = 10 x - 5
3 y = 2 x - 1
so
10 x - 5 = 2 x - 1
8 x = 4
x = 1/2
then
3 y = 2 (1/2) -1
3 y = 1 - 1
y = 0
so (1/2 , 0) is where the two lines intersect
Sure, I can help you with that.
To solve this system of equations, you can use either the substitution method or the elimination method.
Let's use the substitution method to solve the system.
First, let's solve the first equation for x:
10x - 5 = 3y
Add 5 to both sides of the equation:
10x = 3y + 5
Divide both sides by 10:
x = (3y + 5) / 10
Now substitute the value of x in the second equation:
2((3y + 5) / 10) - 3y = 1
Now, let's simplify:
(6y + 10) / 10 - 3y = 1
Multiply both sides by 10 to get rid of the fraction:
6y + 10 - 30y = 10
Combine like terms:
-24y + 10 = 10
Subtract 10 from both sides:
-24y = 0
Divide both sides by -24:
y = 0
Now substitute the value of y back into the first equation:
10x - 5 = 3(0)
Simplify:
10x - 5 = 0
Add 5 to both sides:
10x = 5
Divide both sides by 10:
x = 0.5
So the solution to the system of equations is x = 0.5 and y = 0.