Please help me with ONE small question PLEASE...?

Question

This is in Introductory Calculus Year/grade 11.
The intersection of two lines.

Question: find the intersection of each pair of lines. Remember to write it in the form of y=mx+c first

1) x+4y=5 and x-2y=-1

note: I was able to put them into the form y=mx+c but I ended up with on the 1st equation: 4y=-x-5. Do I have to get the y by itself for it to be in the form y=mx+c? If so do I just divide both sides by 4 and have the two on the right hand side of the equals side as fractions? Won't that make finding the point of intersection quite difficult when I do the next equation to find x and then y? I really am not sure what to do. Please help me and please explain it to me also if possible.

Please tell me if you're not entirely sure of how to do this. If this helps, the final answer for the point of intersection (where the two lines meet together) is:

x=1, y=1 Point of intersection: (1,1).

Sorry, typo. It should be: Find the intersection of this (not these) pair of lines

Answer 1

x + 4y = 5
=> y = 5/4 - x/4
x - 2y = -1
=> y = 1/2 + x/2
=> 5/4 - x/4 = 1/2+x/2
=>5/4-1/2 = x/4+x/2
=>3/4 = 3x/4
=> x = 1
=> 1 + 4y = 5
=> y=1
intersection is at (1,1)

Answer 2

Yes, you do need to divide both sides of 4y = -x +5 (note that it's +5, not -5) by 4. The expression "y = mx + b" implies that there is no coefficient in front of the y.

As for making things difficult, for finding the point of intersection, I don't think it will. Once you have both lines in the form y =mx+b, you can just set the right side of both expressions equal to each other and solve for "x". If you want to get rid of the fractions, you can just multiply by the LCD:

y = -x/4 + 5/4
y = x/2 + 1/2

-x/4 + 5/4 = x/2 + 1/2
4(-x/4 + 5/4) = 4(x/2 + 1/2)
-x + 5 = 2x + 2
3 = 3x
x = 1

Now that you know that the x value of the point of interseciton is 1, plug this back into either of the original equations to find y. The fact that you rewrote them as y=mx+b makes this step easier.

Answer 3

If you're asked to write it in the form y=mx+c, then you have to make sure that it is just y, not 4y. So yes, you have to divide both sides by 4 to get y = -(1/4)x + 5/4. The other equation becomes y = (1/2)x + 1/2

To get the point of intersection, you find a point which fits both equations. So the two y's will be equal and the two x's will be equal. So you just make the two equations equal:

y = -(1/4)x + 5/4 = (1/2)x + 1/2

And then solve it.

Answer 4

Yes, divide both sides by 4 to get into the standard line form. That's your homework.

Next, definition of an intersection between two lines is a point (x,y) that satisfies both equations.

Starting with the standard form, you have y on the left side of both equations. And you know y = y (especially at the intersect).
So you can equate the right side of both equations.

m1 x + c1 = m2 x + c2

Now you can solve for x in this equation. Once you have that x, you can get y by putting the solved x value into either one of the equations.

Note typo: x+4y = 5 is rewritten as 4y = -x +5. Then you divide by 4.

good luck

Answer 5

You have two equations:

x + 4y = 5 ------------------------------------------- (1)
x – 2y = -1 ------------------------------------------- (2)

You have to solve this equation one by one.

We do the first equation first:-

x + 4y = 5

Rearrange this as: - y = mx + c

So: 4y = -x + 5
y = (-x + 5)/4
= -x/4 + 5/4

Therefore: y = (-1/4)x + (5/4)

The intersection is at 5/4 on the y-axis

From second equation:-

x – 2y = -1

Again rearrange into y = mx + c form

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

The intersection is at 1/2 on the y-axis

Answer 6

...nothing like making a problem more difficult than it has to be...(You can solve it more easily if it is NOT in slope-intercept form)...no wonder kids come to hate math!
y = -(1/4)x + 5/4
y = (1/2)x + 1/2
(1/2)x + 1/2 = - (1/4)x + 5/4
(1/2)x + (1/4)x = 5/4 - 1/2
2x + x = 5 - 2
3x = 3
x = 1
y = 1/2 + 1/2 = 1
point of intersection = (1,1)

Note the difference if you are not required to use slope-intercept form:
x + 4y = 5
x - 2y = -1
subtracting:
6y = 6
y = 1
substituting back,
x + 4*1 = 5
x = 1

Answer 7

once you have them as y=-1/4 x + 5/4 and y= 1/2 x + 1/2, set them equal to each other to solve for x:

-1/4 x + 5/4 = 1/2 x + 1/2

move anything attached to an x to one side, all else to the other side:

3/4 = 3/4x

divide each by 3/4:

1=x

then sub that back into either of the equations:

1/2 x +1/2=y

1/2 (1) +1/2 =y

1/2 + 1/2 = y

1=y

So the point is (1,1)

Answer 8

Consider the following two equations:-

x + 4y = 5 (x by 1)
x - 2y = - 1 (x by 2)

x + 4y = 5
2x - 4y = -2

Adding gives 3x = 3
x = 1

Now x + 4y = 5 therefore
1 + 4y = 5
4y = 4
y = 1

Point of intersection is (1, 1)

Answer 9

Find the point of intersection of the two lines.

x + 4y = 5
x - 2y = -1

Subtract the second equation from the first.

6y = 6
y = 1

Now plug the y value back into either equation to solve for x. It doesn't matter which equation. Let's use the first one.

x + 4y = 5
x + 4*1 = 5
x + 4 = 5
x = 1

So your point of intersection (x,y) = (1,1).

Answer 10

you can try to get y =mx+c or in this case it is easier to go for x = my+c .....
x+4y=5 becomes x=5-4y and
x-2y=-1 becomes x=2y-1
now we can say x=x so 5-4y=2y-1
move all the y to one side 5+1=2y+4y so 6=6y and simplify to y=1
go back to equation x+4y=5 and put y=1 x+4(1)=5 x=5-4=1
so the intersection point is (1,1)

if you do it your way then
x+4y=5 becomes 4y=5-x y=(5-x)/4
x-2y=-1 becomes -2y=-1-x y = (1+x)/2
put y=y together and (5-x)/4=(1+x)/2 and cross multiply
2(5-x) = 4(1+x) 10-2x=4+4x 10-4=2x+4x adn 6 = 6x x=1

I hope this helps