Substitution Method for Solving Systems of Equations

Now, if you are in a higher math class, you can also look at the graphs of these two equations and see that they are parallel lines that never intersect, meaning no solutions. In summary, the equations 5a = 5 - b and 5a = 3 - b have no solutions. This can be determined by solving for b in each equation and seeing that they are parallel lines with the same slope, or by graphing the equations and seeing that they do not intersect.
  • #1
rcmango
234
0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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.
 
  • #5
I see now, there are no solutions, anyway you work it out. Thankyou.
 
  • #6
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.
 

FAQ: Substitution Method for Solving Systems of Equations

What is substitution in algebra?

Substitution in algebra is the process of replacing a variable with a number or expression in an algebraic equation.

Why do we use substitution in algebra?

Substitution is used in algebra to simplify and solve equations by replacing unknown variables with known values.

How do we perform substitution in algebra?

To perform substitution in algebra, first identify the variable that needs to be replaced. Then, substitute the value or expression in place of the variable and simplify the resulting equation.

What are the benefits of using substitution in algebra?

Using substitution in algebra can help us solve complex equations and systems of equations. It also allows us to find the value of a variable in terms of other known quantities.

Can substitution be used in all types of algebraic equations?

Yes, substitution can be used in all types of algebraic equations, including linear, quadratic, and exponential equations. It is a versatile and powerful tool in algebraic problem-solving.

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