Suppose I tell you that
x36=1536x36=1536 size 12{ { {x} over {"36"} } = { {"15"} over {"36"} } } {}. What are all the values
x
x can take that make this statement true?

OK, that was easy, wasn’t it? So the moral of that story is: rational equations are easy to solve, if they have a common denominator. If they don’t, of course, you just get one!

Now suppose I tell you that
x18=1536x18=1536 size 12{ { {x} over {"18"} } = { {"15"} over {"36"} } } {}. What are all the values
xx can take that make this statement true?

*Hey,
x
x came out being a fraction. Can he do that?*

Umm, yeah.

OK, one more pretty easy one.

x
2
+
2
21
=
9
7
x
2
+
2
21
=
9
7
size 12{ { {x rSup { size 8{2} } +2} over {"21"} } = { {9} over {7} } } {}

Did you get only one answer? Then look again—this one has two!

Once you are that far, you’ve got the general idea—get a common denominator, and then set the numerators equal. So let’s really get into it now…

x
+
2
x
+
3
=
x
+
5
x
+
4
x
+
2
x
+
3
=
x
+
5
x
+
4
size 12{ { {x+2} over {x+3} } = { {x+5} over {x+4} } } {}

2x
+
6
x
+
3
=
x
+
5
2x
+
7
2x
+
6
x
+
3
=
x
+
5
2x
+
7
size 12{ { {2x+6} over {x+3} } = { {x+5} over {2x+7} } } {}

x+32x−3=x−5x−4x+32x−3 size 12{ { {x+3} over {2x - 3} } } {}=x−5x−4 size 12{ { {x - 5} over {x - 4} } } {}

**a. **Solve. You should end up with two answers. **b. **Check both answers by plugging them into the original equation.

Comments:"This is the "main" book in Kenny Felder's "Advanced Algebra II" series. This text was created with a focus on 'doing' and 'understanding' algebra concepts rather than simply hearing about them in […]"