Suppose that |x-2|<0.01 and |y-3|<0.04 .Use the Triangle Inequality to show
that |(x+y)-5|<0.05 represent your solution on a number line.
By the Triangle Inequality, we have:
|(x+y)-5| <= |x-2| + |y-3|
Since we know that |x-2|<0.01 and |y-3|<0.04, we can substitute:
|(x+y)-5| <= 0.01 + 0.04
Simplifying:
|(x+y)-5| < 0.05
This shows that if |x-2|<0.01 and |y-3|<0.04, then |(x+y)-5|<0.05. On a number line, this means that the solutions lie within a shaded interval of length 0.05, centered at 5.
To show that |(x+y)-5|<0.05 using the Triangle Inequality, we can start by applying the Triangle Inequality to the expression |x-2|<0.01:
|x-2| < 0.01
|x-2+2| < 0.01 + 2
|x| < 2.01
Similarly, we can apply the Triangle Inequality to the expression |y-3|<0.04:
|y-3| < 0.04
|y-3+3| < 0.04 + 3
|y| < 3.04
Now, let's substitute these results into the expression |(x+y)-5| and use the Triangle Inequality:
|(x+y)-5| = |x+y-5|
By the Triangle Inequality, we know that |a + b| ≤ |a| + |b| for any real numbers a and b. Therefore:
|x+y-5| ≤ |x| + |y-5|
Using the results we found earlier:
|x+y-5| ≤ 2.01 + 3.04
Simplifying:
|x+y-5| ≤ 5.05
Since 5.05 is greater than 0.05, we can say that |(x+y)-5|<0.05 is true.