SOLUTION: Solving a System of Equations. Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in ordered-pair form. 3/4x+1/2y=5 -1

Algebra ->  Linear-equations -> SOLUTION: Solving a System of Equations. Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in ordered-pair form. 3/4x+1/2y=5 -1      Log On


   

Question 1128374: Solving a System of Equations. Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in ordered-pair form.
3/4x+1/2y=5
-1/4x-3/2y=1
Found 3 solutions by MathLover1, ikleyn, MathTherapy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

%283%2F4%29x%2B%281%2F2%29y=5
%28-1%2F4%29x-%283%2F2%29y=1

Solved by pluggable solver: Solve the System of Equations by Graphing


Let's look at the first equation %283%2F4%29x%2B%281%2F2%29y=5



4%28%283%2F4%29x%2B%281%2F2%29y%29=4%285%29 Multiply both sides of the first equation by the LCD 4



3x%2B2y=20 Distribute



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Let's look at the second equation %281%2F4%29x%2B%283%2F2%29y=1


4%28%281%2F4%29x%2B%283%2F2%29y%29=4%281%29 Multiply both sides of the second equation by the LCD 4



1x%2B6y=4 Distribute



---------




So our new system of equations is:


3x%2B2y=20

1x%2B6y=4





In order to graph these equations, we need to solve for y for each equation.




So let's solve for y on the first equation


3x%2B2y=20 Start with the given equation



2y=20-3x Subtract 3+x from both sides



2y=-3x%2B20 Rearrange the equation



y=%28-3x%2B20%29%2F%282%29 Divide both sides by 2



y=%28-3%2F2%29x%2B%2820%29%2F%282%29 Break up the fraction



y=%28-3%2F2%29x%2B10 Reduce



Now lets graph y=%28-3%2F2%29x%2B10 (note: if you need help with graphing, check out this solver)



+graph%28+600%2C+600%2C+-10%2C+10%2C+-10%2C+10%2C+%28-3%2F2%29x%2B10%29+ Graph of y=%28-3%2F2%29x%2B10




So let's solve for y on the second equation


1x%2B6y=4 Start with the given equation



6y=4-x Subtract +x from both sides



6y=-x%2B4 Rearrange the equation



y=%28-x%2B4%29%2F%286%29 Divide both sides by 6



y=%28-1%2F6%29x%2B%284%29%2F%286%29 Break up the fraction



y=%28-1%2F6%29x%2B2%2F3 Reduce





Now lets add the graph of y=%28-1%2F6%29x%2B2%2F3 to our first plot to get:


Graph of y=%28-3%2F2%29x%2B10(red) and y=%28-1%2F6%29x%2B2%2F3(green)


From the graph, we can see that the two lines intersect at the point (7,-1%2F2) (note: you might have to adjust the window to see the intersection)

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Your system can be read in this way


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


or in this way


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


depending on how to put PARENTHESES into equations.


So, the way you presented the problem is AMBIGOUS.


To avoid ambiguity, YOU must put parentheses in your formula BEFORE POSTING IT TO THE FORUM.



Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Solving a System of Equations. Solve the system, or show that it has no solution. If the system has infinitely many solutions, express them in ordered-pair form.
3/4x+1/2y=5
-1/4x-3/2y=1
That's the most inefficient, most time-consuming, and most error-prone method I've ever seen one use to solve a system of equations, 

if I ever saw one. It is so inefficient and CONFUSING to the point where the woman who solved it got WRONG solutions. Her solutions 

DO NOT SATISFY THE EQUATIONS, so IGNORE HER ENTIRE SOLUTION. Plus, why would someone even try to solve a fractional system by graphing? RIDICULOUS!



I assume the system is:            matrix%281%2C3%2C+%283%2F4%29x+%2B+%281%2F2%29y%2C+%22=%22%2C+5%29 ------ eq (i)

                                   matrix%281%2C3%2C+%28-+1%2F4%29x+-+%283%2F2%29y%2C+%22=%22%2C+1%29 ---- eq (ii) 

Just multiply eq (ii) by 3 to get: matrix%281%2C3%2C+%28-+3%2F4%29x+-+%289%2F2%29y%2C+%22=%22%2C+3%29 ---- eq (iii)

Adding eqs (i) & (iii) results in: matrix%281%2C3%2C+%281%2F2%29y+-+%289%2F2%29y%2C+%22=%22%2C+8%29

matrix%281%2C3%2C+%28-+8%2F2%29y%2C+%22=%22%2C+8%29		

- 4y = 8

highlight_green%28matrix%281%2C3%2C+y%2C+%22=%22%2C+highlight%28-+2%29%29%29%29		

Substitute - 2 for y in any of the original equations and you should get: highlight_green%28matrix%281%2C3%2C+x%2C+%22=%22%2C+8%29%29

That's all! Nothing more, nothing less!!