SOLUTION: Graph 3x+4y=7 and 5x+2y=12. I know how to do this. The next part of the problem asks: Find the exact coordinates of their point of intersection, and check your answer by substituti

Algebra ->  Graphs -> SOLUTION: Graph 3x+4y=7 and 5x+2y=12. I know how to do this. The next part of the problem asks: Find the exact coordinates of their point of intersection, and check your answer by substituti      Log On


   

Question 123074: Graph 3x+4y=7 and 5x+2y=12. I know how to do this. The next part of the problem asks: Find the exact coordinates of their point of intersection, and check your answer by substitution in the given equations. How do you go about doing this ?
Found 2 solutions by jim_thompson5910, rapaljer:
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Note: It will be very hard to find the exact coordinates of the intersection with just the graph (it turns out that the coordinates of the intersection are fractions). So to find the exact coordinates of the intersection, we need to use either substitution or elimination. So let's use substitution

Start with the given system of equations:

system%283x%2B4y=7%2C5x%2B2y=12%29




In order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to solve for y.




So let's isolate y in the first equation

3x%2B4y=7 Start with the first equation


4y=7-3x Subtract 3x from both sides


4y=-3x%2B7 Rearrange the equation


y=%28-3x%2B7%29%2F%284%29 Divide both sides by 4


y=%28%28-3%29%2F%284%29%29x%2B%287%29%2F%284%29 Break up the fraction


y=%28-3%2F4%29x%2B7%2F4 Reduce



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

Since y=%28-3%2F4%29x%2B7%2F4, we can now replace each y in the second equation with %28-3%2F4%29x%2B7%2F4 to solve for x



5x%2B2highlight%28%28%28-3%2F4%29x%2B7%2F4%29%29=12 Plug in y=%28-3%2F4%29x%2B7%2F4 into the first equation. In other words, replace each y with %28-3%2F4%29x%2B7%2F4. Notice we've eliminated the y variables. So we now have a simple equation with one unknown.



5x%2B%282%29%28-3%2F4%29x%2B%282%29%287%2F4%29=12 Distribute 2 to %28-3%2F4%29x%2B7%2F4


5x-%286%2F4%29x%2B14%2F4=12 Multiply


%284%29%285x-%286%2F4%29x%2B14%2F4%29=%284%29%2812%29 Multiply both sides by the LCM of 4. This will eliminate the fractions (note: if you need help with finding the LCM, check out this solver)



20x-6x%2B14=48 Distribute and multiply the LCM to each side



14x%2B14=48 Combine like terms on the left side


14x=48-14Subtract 14 from both sides


14x=34 Combine like terms on the right side


x=%2834%29%2F%2814%29 Divide both sides by 14 to isolate x



x=17%2F7 Reduce





-----------------First Answer------------------------------


So the first part of our answer is: x=17%2F7









Since we know that x=17%2F7 we can plug it into the equation y=%28-3%2F4%29x%2B7%2F4 (remember we previously solved for y in the first equation).



y=%28-3%2F4%29x%2B7%2F4 Start with the equation where y was previously isolated.


y=%28-3%2F4%29%2817%2F7%29%2B7%2F4 Plug in x=17%2F7


y=-51%2F28%2B7%2F4 Multiply


y=-1%2F14 Combine like terms and reduce. (note: if you need help with fractions, check out this solver)



-----------------Second Answer------------------------------


So the second part of our answer is: y=-1%2F14









-----------------Summary------------------------------

So our answers are:

x=17%2F7 and y=-1%2F14

which form the point








Now let's graph the two equations (if you need help with graphing, check out this solver)


From the graph, we can see that the two equations intersect at . This visually verifies our answer.




graph of 3x%2B4y=7 (red) and 5x%2B2y=12 (green) and the intersection of the lines (blue circle).





Check:


Start with the given system of equations:

system%283x%2B4y=7%2C5x%2B2y=12%29


Plug in x=17%2F7 and y=-1%2F14:

system%283%2817%2F7%29%2B4%28-1%2F14%29=7%2C5%2817%2F7%29%2B2%28-1%2F14%29=12%29


Multiply

system%2851%2F7-4%2F14=7%2C85%2F7-2%2F14=12%29


Combine like terms

system%287=7%2C12=12%29


Since we have true equalities, this verifies our answer.

Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
3x+4y=7
5x+2y=12

You need to find a way to eliminate one of the variables, by multiplying one or both of the equations by some number(s), and adding them together. In this case, since you have a 4y and a 2y, if you want to make the 4y subtract out, you will need a -4y, so you should multiply the second equation by -2.

3x+4y=7
-10x-4y=-24

-7x = -17
x=+17%2F7

Since this came out to such an ugly answer, I would eliminate the x terms in the same way. You have coefficients of x of 3 and 5. A "common number" would be 15, so multiply the first equation by 5 and the second equation by -3, which will give you a 15x and -15x, which subtract out.

3x+4y=7
5x+2y=12

5(3x+4y)=5(7)
-3(5x+2y)=-3(12)

15x+20y = 35
-15x-6y=-36

14y=-1
y=-1/14

This is REALLY ugly, so check it by substituting into the first equation:
3x+4y=7
3(17/7) +4(-1/14) = 7
51/7 -2/7 = 7
49/7=7

Some instructors require a check in BOTH equations. The second check will be similar to this!

R^2