If you reverse the sense of the equation, you have
0 x= 4
Now you can subtract 4 from each side, getting
0 x− 4 = 0
It’s tempting to think that this is in standard form for a first-degree equation. But the constant by which
you multiply x (called the coefficient of x) is 0, so the variable x becomes meaningless. No matter what real
number you choose for x, you get an absurd result.
Word Problems
Centuries ago, the algebra in this chapter was unknown, even to the best mathematicians.
Problems that seem simple to us were difficult for them. When they encountered word problems
like the ones that follow, they often sought out solutions by making educated guesses until
they “got lucky.” We have a better way.
Problem A
Imagine a number. We add it to half of itself, and then add the result to 1/5 of itself. The final
sum is equal to 51/10. What is the original number?
Solution A
The first step in solving any word problem is to set up an equation representing that problem.
Let’s call our unknown number x.The statement of the problem tells us that if we take x,then
addx/2 to that, and then add x/5 to that, we get 51/10. The equation is
x+x/2+x/5= 51/10
We can multiply through by 10 to get
10 x+ 5 x+ 2 x= 51
Let’s apply the right-hand distributive law for multiplication over addition “in reverse” on the
left side of the equation. That gives us
(10+ 5 + 2)x= 51
which simplifies to
17 x= 51
We can divide through by 17 and get
x= 3
Word Problems 203
