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Home Q & A Board Science-Related Homework Help Mathematics
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rkiecaboose rkiecaboose
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11 years ago
Find the point of intersection of the straight lines with equations:

x + 2y = -5 and 3x - y = 13


Can someone please tell me how I go about achieving the answer to this question, as I have no idea where to start. This is revision before my exam so as much detail as possible please. Thanks.
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#1
11 years ago
substitution method is the easiest. solve for one variable, say x in the first equation getting x = -2y -5
then put that expression in the second equation 3(-2y -5) - y = 13 solve for y
-6y -15 -y = 13
-7y = 28
y = -4
using y plug into equation for x : x = -2(-4) -5 = 8 -5 = 3
wrote...
#2
11 years ago
Well friend all lines are straight Smiling Face with Open Mouth
To start off you can solve for y for both equations...

y = (-1/2)x - (5/2)      and    y = 3x -13

make both equations equal to each other in order to find the value of your x-coordinate....

solve for x and you find that it is equal to 3.

Now plug the 3 into one of your starting equations, I'll use x + 2y = -5 and solve for y

y then equals -4.
So your point of intersection is (3,-4)

Hope this helped Slight Smile
wrote...
#3
11 years ago
The procedure in a nutshell is:
Solve both equations for y:
y1=a1*x+b1
y2=a2*x+b2
Then equate the two RHSs:
a1*x+b1=a2*x+b2
and solve for x:
x=-(b2-b1)/(a2-a1)
Now you have the x-value where the lines intersect.
Take this x-value and plug it into any of the two
equations to get the respective y-value.
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