Find the intersection of two lines. i have the answer just please tell me how it was made
4x-3y-4=0 , 4x+2y+5=0
-5y/-5 = 9/-5
y= 9/5
4x-3(-9/5)-4=0
5(4x+27/5-4=0)
20x+27-20=0
20x+7=0
20x/20=-7/20
x=-7/20
(-7/20, -9/5
If you subtract the 2nd equation from the first, you get
4x-3y-4 = 0
4x+2y+5 = 0
----------------
-5y-9 = 0
now add 9 to both sides:
-5y = 9
now divide by -5:
y = -9/5
Knowing that y = -9/5, just plug it into either of the original equations and then solve for x. The work is shown above. If you have trouble following the steps in logic, you need to do a bunch of these on your own.
To find the intersection point of two lines, you need to solve the system of equations formed by the two lines. In this case, the given lines are:
1) 4x - 3y - 4 = 0
2) 4x + 2y + 5 = 0
To solve this system, we'll use the method of substitution.
First, solve equation 1) for x in terms of y:
4x - 3y - 4 = 0
4x = 3y + 4
x = (3y + 4)/4
Now, substitute this expression for x in equation 2):
4( (3y + 4)/4 ) + 2y + 5 = 0
3y + 4 + 2y + 5 = 0
5y + 9 = 0
5y = -9
y = -9/5
Substituting this value of y into equation 1) gives:
4x - 3(-9/5) - 4 = 0
4x + 27/5 - 4 = 0
4x + 27/5 - 20/5 = 0
4x + 7/5 = 0
4x = -7/5
x = -7/20
Therefore, the intersection point is (-7/20, -9/5).