SOLUTION: Solve the system of linear equations using substitution method: 2x+3y=5 2x+y=3

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Question 90012: Solve the system of linear equations using substitution method:
2x+3y=5
2x+y=3
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Solving a linear system of equations by subsitution


Lets start with the given system of linear equations

2%2Ax%2B3%2Ay=5
2%2Ax%2B1%2Ay=3

Now in order to solve this system by using substitution, we need to solve (or isolate) one variable. I'm going to choose y.

Solve for y for the first equation

3%2Ay=5-2%2AxSubtract 2%2Ax from both sides

y=%285-2%2Ax%29%2F3 Divide both sides by 3.


Which breaks down and reduces to



y=5%2F3-%282%2F3%29%2Ax Now we've fully isolated y

Since y equals 5%2F3-%282%2F3%29%2Ax we can substitute the expression 5%2F3-%282%2F3%29%2Ax into y of the 2nd equation. This will eliminate y so we can solve for x.


2%2Ax%2B1%2Ahighlight%28%285%2F3-%282%2F3%29%2Ax%29%29=3 Replace y with 5%2F3-%282%2F3%29%2Ax. Since this eliminates y, we can now solve for x.

2%2Ax%2B1%2A%285%2F3%29%2B1%28-2%2F3%29x=3 Distribute 1 to 5%2F3-%282%2F3%29%2Ax

2%2Ax%2B5%2F3-%282%2F3%29%2Ax=3 Multiply



2%2Ax%2B5%2F3-%282%2F3%29%2Ax=3 Reduce any fractions

2%2Ax-%282%2F3%29%2Ax=3-5%2F3 Subtract 5%2F3 from both sides


2%2Ax-%282%2F3%29%2Ax=9%2F3-5%2F3 Make 3 into a fraction with a denominator of 3


2%2Ax-%282%2F3%29%2Ax=4%2F3 Combine the terms on the right side



%286%2F3%29%2Ax-%282%2F3%29x=4%2F3 Make 2 into a fraction with a denominator of 3

%284%2F3%29%2Ax=4%2F3 Now combine the terms on the left side.


cross%28%283%2F4%29%284%2F3%29%29x=%284%2F3%29%283%2F4%29 Multiply both sides by 3%2F4. This will cancel out 4%2F3 and isolate x

So when we multiply 4%2F3 and 3%2F4 (and simplify) we get



x=1 <---------------------------------One answer

Now that we know that x=1, lets substitute that in for x to solve for y

2%281%29%2B1%2Ay=3 Plug in x=1 into the 2nd equation

2%2B1%2Ay=3 Multiply

1%2Ay=3-2Subtract 2 from both sides

1%2Ay=1 Combine the terms on the right side

cross%28%281%2F1%29%281%29%29%2Ay=%281%2F1%29%281%2F1%29 Multiply both sides by 1%2F1. This will cancel out 1 on the left side.

y=1%2F1 Multiply the terms on the right side


y=1 Reduce


So this is the other answer


y=1<---------------------------------Other answer


So our solution is

x=1 and y=1

which can also look like

(1,1)

Notice if we graph the equations (if you need help with graphing, check out this solver)

2%2Ax%2B3%2Ay=5
2%2Ax%2B1%2Ay=3

we get


graph of 2%2Ax%2B3%2Ay=5 (red) and 2%2Ax%2B1%2Ay=3 (green) (hint: you may have to solve for y to graph these) intersecting at the blue circle.


and we can see that the two equations intersect at (1,1). This verifies our answer.


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Check:

Plug in (1,1) into the system of equations


Let x=1 and y=1. Now plug those values into the equation 2%2Ax%2B3%2Ay=5

2%2A%281%29%2B3%2A%281%29=5 Plug in x=1 and y=1


2%2B3=5 Multiply


5=5 Add


5=5 Reduce. Since this equation is true the solution works.


So the solution (1,1) satisfies 2%2Ax%2B3%2Ay=5



Let x=1 and y=1. Now plug those values into the equation 2%2Ax%2B1%2Ay=3

2%2A%281%29%2B1%2A%281%29=3 Plug in x=1 and y=1


2%2B1=3 Multiply


3=3 Add


3=3 Reduce. Since this equation is true the solution works.


So the solution (1,1) satisfies 2%2Ax%2B1%2Ay=3


Since the solution (1,1) satisfies the system of equations


2%2Ax%2B3%2Ay=5
2%2Ax%2B1%2Ay=3


this verifies our answer.