Free Math Worksheets

- Substitute each value of \(x\), using parentheses around negative values. - Subtracting a negative number is the same as adding its opposite. - Compare the three results. - Think about what happens to a difference when the number being subtracted becomes smaller.

1. For \(x = 2\), \(-4.5 - 2 = -6.5\). 2. For \(x = -2\), \(-4.5 - (-2) = -4.5 + 2 = -2.5\). 3. For \(x = -4.5\), \(-4.5 - (-4.5) = 0\). 4. As \(x\) decreases from \(2\) to \(-2\) to \(-4.5\), the expression increases from \(-6.5\) to \(-2.5\) to \(0\). 5. When \(x\) becomes smaller, its opposite \(-x\) becomes larger, so \(-4.5 - x\) becomes larger.

a) For \(x = 2\), the value is \(-6.5\); for \(x = -2\), it is \(-2.5\); for \(x = -4.5\), it is \(0\). b) The value of the expression increases as \(x\) decreases because \(-x\) increases.

- Rewrite every term using tenths. - Express one variable in terms of the other, or test small positive whole-number values systematically. - Substitute each proposed pair back into the original equation. - Check whether more than one positive whole-number pair satisfies the equation.

1. Rewrite \(0.7\) as \(\frac{7}{10}\) and \(\frac{y}{5}\) as \(\frac{2y}{10}\). The equation becomes \(x+2y=7\). 2. For positive whole numbers, possible values are \((x, y)=(5, 1)\), \((3, 2)\), and \((1, 3)\). 3. Substitution verifies each pair: \(\frac{5}{10}+\frac{1}{5}=0.7\), \(\frac{3}{10}+\frac{2}{5}=0.7\), and \(\frac{1}{10}+\frac{3}{5}=0.7\).

Possible pairs are \((5, 1)\), \((3, 2)\), and \((1, 3)\). Any two of these satisfy the requirement.

- Substitute the given values of \(x\) into the expression. - Rewrite the division as one fraction with \(x\) in the denominator. - Think about what happens to a positive fraction when its denominator increases.

1. For a), when \(x=2\), \(T=\frac{4}{5}\div2=\frac{2}{5}\). When \(x=8\), \(T=\frac{4}{5}\div8=\frac{1}{10}\). 2. For b), \(T=\frac{4}{5x}\). Set \(\frac{4}{5x}=\frac{1}{10}\). Since \(\frac{1}{10}=\frac{4}{40}\), \(5x=40\), so \(x=8\). 3. For c), as \(x\) increases, the denominator \(5x\) increases while the numerator stays \(4\). Therefore, \(T\) decreases and approaches \(0\).

a) When \(x=2\), \(T=\frac{2}{5}\); when \(x=8\), \(T=\frac{1}{10}\). b) \(x=8\) c) \(T\) decreases as \(x\) increases and approaches \(0\).