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Select for a and for b values such that the equation \(\frac{\mathrm{x}}{\mathrm{a}} = \mathrm{x} - \mathrm{b}\) has more than one solution for \(\mathrm{x}\).
Make only two selections, one in each column.
-2
-1
0
1
2
We need to find values of a and b such that the equation \(\mathrm{x/a = x - b}\) has more than one solution for x.
For this algebraic problem, let's use equation format to track our transformations:
Original: \(\mathrm{x/a = x - b}\)
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Goal: Find a and b that give MORE THAN ONE solution
For a linear equation in one variable, we can have:
"More than one solution" means infinitely many solutions, which happens when our equation becomes an identity.
Let's manipulate \(\mathrm{x/a = x - b}\) algebraically.
Multiplying both sides by a (assuming \(\mathrm{a ≠ 0}\)):
\(\mathrm{x = a(x - b)}\)
\(\mathrm{x = ax - ab}\)
\(\mathrm{x - ax = -ab}\)
\(\mathrm{x(1 - a) = -ab}\)
If \(\mathrm{a ≠ 1}\), we can divide by \(\mathrm{(1 - a)}\):
\(\mathrm{x = -ab/(1 - a)}\)
This gives exactly ONE solution - not what we want.
If \(\mathrm{a = 1}\), then:
\(\mathrm{x(1 - 1) = -1 × b}\)
\(\mathrm{x × 0 = -b}\)
\(\mathrm{0 = -b}\)
This means b must equal 0.
When \(\mathrm{a = 1}\) and \(\mathrm{b = 0}\):
\(\mathrm{x/1 = x - 0}\)
\(\mathrm{x = x}\)
This is an identity! It's true for ALL values of x, giving us infinitely many solutions.
From [-2, -1, 0, 1, 2]:
Both values are available in our choices.
This makes the equation \(\mathrm{x/1 = x - 0}\), which simplifies to \(\mathrm{x = x}\), an identity true for all x values, giving us infinitely many solutions.