SOLUTION: I am totaly lost on solving orderd pair, I know how to graph once i find the answer but have no idea where to begin. My teacher said you could plug in any numbers but doesnt it sti
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Question 798767: I am totaly lost on solving orderd pair, I know how to graph once i find the answer but have no idea where to begin. My teacher said you could plug in any numbers but doesnt it still have to equal the sum.
6y=4x-14
Found 3 solutions by mananth, josgarithmetic, MathTherapy:
Answer by mananth(16946) (Show Source): You can put this solution on YOUR website!
6y=4x-14
/2
3y=2x-7
/3
y=(2/3)x-(7/3)
Plug
x=3 , y=(-1/3), (3,(-1/3))
x=6, y=(5/3) ,(6,(5/3))
x=-3, (-13/3) (-3,(-13/3))
Rather difficult
plot these points
Answer by josgarithmetic(39620) (Show Source): You can put this solution on YOUR website!
Your teacher is correct. You have a linear equation in two variables. You can plug in any value for one variable and solve for the value of the other variable.
"Where to begin..." -------
PICK ANY VALUE FOR ONE OF THE VARIABLES!
SOLVE FOR THE OTHER VARIABLE!
Your given equation is . I suggest you simplify this, seeing that all coefficients are EVEN numbers.
You will find convenient solving for either variable, NOW!
In this formula, choose a value for x and compute the corresponding value for y. Doing so gives you a point, an ordered pair, (x,y).
Answer by MathTherapy(10552) (Show Source): You can put this solution on YOUR website!
I am totaly lost on solving orderd pair, I know how to graph once i find the answer but have no idea where to begin. My teacher said you could plug in any numbers but doesnt it still have to equal the sum.
6y=4x-14
2(3y) = 2(2x - 7) ----- Factoring out GCF, 2
3y = 2x - 7
To find ordered pairs without being too "messy," use the following x-values and determine the corresponding y-value:
x = 5
x = 11
x = 17
x = 23
Using the ordered pairs, plot the graph of the linear equation.
Send comments, “thank-yous,” and inquiries to “D” at MathMadEzy@aol.com. Further help is available, online or in-person, for a fee, obviously.
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