Need Help! Find the point of intersection of the given pairs of straight lines.?

Question

Problem One: 2x + 3y = 1
3x + y = 5

Problem Two: 2/3x + 5y = 12
x + 1/2y = 4

I just want to know the algebraic steps, to solve the equation. I'm totally confused here. Help would be greatly appreciated!

Answer 1

Simple. Notice that with a little manipulation in Problem One, 2x +3y -1 = 0. And, 3x + y - 5 = 0.
Therefore, since both of those equations are equal to 0, that means 2x + 3y - 1 = 3x + y - 5.
2x + 3y = 3x + y - 4
2x + 2y = 3x - 4
2y = x - 4 or x = 2y + 4

So, we've found a formula to give us the value of x in the intersection, relative to the value of y in the same intersection.

Now we can substitute the relative value of x (which we know is 2y + 4) into all occurrences of x in either of the original equations, to find the actual value of y:

2x + 3y = 1
So, 2(2y + 4) + 3y = 1
4y + 8 + 3y = 1
7y + 8 = 1
7y = -7
y = -1

Now, substitute -1 for y in either of original equations to find the actual value of x in the intersection:

3x + y = 5
3x -1 = 5
3x = 6
x =2

Therefore, the intersection of the equations in Problem One occurs at (2, -1)

The same steps can be used to find that the intersection in Problem Two would be (3, 2).

Answer 2

1)

2x + 3y = 1
3x + y = 5

multiply equation 2 by 3
2x + 3y = 1
9x + 3y = 15

subtract upward to make things easier
7x = 14
x = 2

plug the x value into any of the equations and solve for y
2(2) + 3y = 1
3y = -3
y = -1

Answer (2, -1)

2)

2/3 x + 5y = 12
x + 1/2 y = 4

Multiply equation 1 by 3/2

x + 15/2 y = 18
x + 1/2 y = 4

subtract downward
7y = 14
y = 2

plug y=2 into any equation
x + 1/2 y = 4
x + 1 = 4
x = 3

Answer (3,2)

Answer 3

First one. From eq1 you get that x = (1-3y)/2. If you replace this in eq2 you get that 3(1-3y)/2+y=5, from where y=-1 and then x=2. So the point of intersection is (2,-1). Same procedure for the second problem.

Answer 4

2x + 3y = 1
3x + y = 5

3x + y = 5
y = -3x + 5

2x + 3(-3x + 5) = 1
2x - 9x + 15 = 1
-7x + 15 = 1
-7x = -14
x = 2

y = -3(2) + 5
y = -6 + 5
y = -1

ANS : (2,-1)

--------------------------------------------

(2/3)x + 5y = 12
x + (1/2)y = 4

x + (1/2)y = 4
x = (-1/2)y + 4

(2/3)((-1/2)y + 4) + 5y = 12
(-2/6)y + (8/3) + 5y = 12
(-1/3)y + (8/3) + 5y = 12
-y + 8 + 15y = 36
14y + 8 = 36
14y = 28
y = 2

x = (-1/2)(2) + 4
x = -1 + 4
x = 3

ANS : (2,3)

Answer 5

2x+3y=1
3y=1-2x
y=(1-2x)/3

3x+y=5
y=5-3x

so (1-2x)/3=5-3x
multiply both sides by 3
1-2x=15-9x
Add 9x to both sides
1+9x-2x=15
1+7x=15
subtract 1 both sides
7x=14
divide both sides by 7
x=2
Plug x=2 into y=5-3x
so y=5-3*2
y=5-6
y=-1
So the point of intersection is (2,-1)


Now do the other one by yourself

Answer 6

first of all you want y by itself then you can graph it. So first flip x oer to the other side and it becomes negative so then you rewrite the equation with x being negative plus whatever number was on the other side. Now the number thats in front of y you take that and divided each varible by that leaving y by itself put the equation into your calculator or draw a graph

Answer 7

You need to get either the x or the y to cancel out first of all, then combine and solve for one letter. Once you have solved one, then plug that value in and you can solve for the other

Answer 8

Try this . . . . . . . .

3x(32ylp/q6uv9
________________

2cp(44/hht4) = 2/.>>^&**(@@:>) :>) :>) += 7.3804 to the 10th power.

conclusion, you need to start paying more attention in class.

Darryl S.