Let's examine the given linear function: \(y = 10 + 2x\).
To determine whether \(y > x\) for all values of \(x\), we need to compare this equation with another expression
involving only \(x\). We can create an inequality and solve it:
\[ y - x > 0 \]
Substitute the given function into our new inequality using \(y = 10 + 2x\):
\[ (10 + 2x) - x > 0 \]
Simplify this to get a simpler expression for comparison:
\[ 10 + x > 0 \]
or equivalently,
\[ x > -10. \]
From the simplified form, we see that as long as \(x > -10\), then:
1. If you pick any value of \(x\) greater than -10 (e.g., if \(x = 2\)), substituting this back into our original
function gives:
\[ y = 10 + 2(2) = 14, \]
which is indeed larger (\(y > x\)).
However, for values less or equal to -10,
- If you pick a value such as \(x = -11\), substituting it back into the original function gives:
\[ y = 10 + 2(-11) = 10 - 22 = -12. \]
Comparing this with our chosen x-value, we see that for an extremely negative number,
\[ -12 < -11,\]
which contradicts \(y > x\).
Thus while the equation tends to hold true in most cases of practical interest (i.e., positive values), it fails
at extreme or very low numerical ranges.
So a counterexample can be shown as:
Let’s take \(x = -10\):
\[ y = 10 + 2(-10) \]
\[ y = 10 - 20 \]
\[ y = -10, \]
Thus,
\[y (-10)\] is not greater than x (-10), thus disproving the statement that "for all values of \(x", then $y > x$.
The counterexample when taking very low value shows it can be validly falsified.