a) To show that the two expressions represent equal numbers when $x=10,$ we can simply substitute $x=10$ into both expressions and compare the results.
For the first expression, we have $1.$
For the second expression, we have $\frac{2(10)+3}{2(10)+3}.$ Simplifying the numerator and denominator separately, we get $\frac{20+3}{20+3} = \frac{23}{23}.$ Since any nonzero number divided by itself equals $1,$ we conclude that $\frac{23}{23} = 1.$
Therefore, when $x=10,$ both expressions are equal to $1.$
b) To show that the two expressions do not represent equal numbers when $x=-\dfrac32,$ we can substitute $x=-\dfrac32$ into both expressions and compare the results.
For the first expression, we have $1.$
For the second expression, we have $\frac{2\left(-\frac32\right)+3}{2\left(-\frac32\right)+3}.$ Simplifying the numerator and denominator separately, we get $\frac{-3+3}{-3+3} = \frac{0}{0}.$
The expression $\frac{0}{0}$ is undefined because division by zero is not allowed in mathematics.
Therefore, when $x=-\dfrac32,$ the first expression is equal to $1$ but the second expression is undefined, so they do not represent equal numbers.
c) To show that the two expressions represent equal numbers for all $x$ other than $-\dfrac32,$ we can simplify the second expression and check if it simplifies to $1$ for all $x$ other than $-\dfrac32.$
The second expression is $\frac{2x+3}{2x+3}.$ Since the numerator and denominator of this fraction are equal, except for $x = -\dfrac32,$ it follows that the fraction is equal to $1$ for all $x$ other than $-\dfrac32.$
Therefore, the two expressions represent equal numbers for all $x$ other than $-\dfrac32.$
Consider the two expressions $1$ and $\frac{2x+3}{2x+3}.$
a) Show that the two expressions represent equal numbers when $x=10.$
b) Explain why these two expressions do not represent equal numbers when $x=-\dfrac32.$
c) Show that these two expressions represent equal numbers for all $x$ other than $-\dfrac32.$
[i]In parts (a) and (c), begin by explaining what your strategy for solving will be.[/i]
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