Substitution Method for Solving Systems of Equations

AI Thread Summary
The discussion centers on solving the system of equations 5a = 5 - b and 5a = 3 - b using the substitution method. Initially, a solution set of (1/2, 5/2) was proposed, but further analysis revealed that both equations yield the same slope of -5, indicating that the lines are parallel. This parallelism means there is no point of intersection, leading to the conclusion that the system of equations has no solution. The incorrect assumption of a solution highlights the importance of analyzing the slopes when dealing with linear equations. Ultimately, the discussion emphasizes the necessity of verifying solutions in systems of equations.
rcmango
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Homework Statement



5a = 5 - b
5a = 3 - b



Homework Equations





The Attempt at a Solution



I got the solution set to be 1/2, 5/2

i used substitution for substituting a into b of the second equation, just like they were x's and y's just used a's and b's there is no difference correct?

a = 1- b/5 from top equation.

then 5(1-b/5) = 3 - b

==5/2

then for 1/2 I substituted the b= 5/2 back into one of the original equations.
Does this look correct? Thanks alot.
 
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Hi rcmango!

You had 5(1-b/5) = 3 - b
If you get rid of the parentheses, you get 5 - b = 3 - b
Can you solve b from that expression?
Or what else can you deduce from that?
 
rcmango said:
5a = 5 - b
5a = 3 - b

Please allow me to translate these equations into a "word problem":

Five apples cost $5 minus a certain bit of change.
Five apples cost $3 minus the same bit.

Does this look as if there could be any fixed price for apples that would make sense in both statements?
 
5a = 5 - b
5a = 3 - b

First equation solved for b: b=5-5a
Substitute into second equation: 5a=3-b,
5a=3-(5-5a)
5a=3-5+5a
add -5a to both sides: 0=-2
Incorrect statement. No solution.

You can try a different way to what happens.
 
I see now, there are no solutions, anyway you work it out. Thankyou.
 
Assuming you are in Introductory Algebra like in high school, another thing you can try is to solve each equation for b. Look at the slope, as if a is the horizontal axis and b is the vertical axis. If the slopes are equal, the lines for each equation are parallel and therefore will not intersect, meaning no point in common, meaning the system of equations has no solution.

b=-5a+5
b=-5a+3

Slope of both equations is -5, so the lines are parallel.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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