Substitution Method for Solving Systems of Equations

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Homework Help Overview

The problem involves a system of equations represented by 5a = 5 - b and 5a = 3 - b, where participants are exploring the substitution method for finding solutions.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the substitution of variables and the implications of the resulting equations. There are attempts to interpret the equations in a real-world context and to analyze the relationships between the variables.

Discussion Status

Some participants have identified that the equations lead to a contradiction, suggesting there is no solution. Others are exploring the implications of the slopes of the equations to confirm the absence of intersections.

Contextual Notes

There is a focus on understanding the nature of the equations and their graphical representation, with some participants questioning the setup and assumptions regarding the relationships between a and b.

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 a lot.
 
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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.
 

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