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task d: solve it! determine the value of the variable that makes each e…

Question

task d: solve it!
determine the value of the variable that makes each equation true.

  1. $-11.2 - a = 0$
  2. $-1.1b = -5.5$
  3. $2c = -7$
  4. $\frac{d}{-6} = 1$
  5. $8.3 + f = 5.2$
  6. $g + (-2) = -6.5$

for two new expressions, $c - d$ and $c \cdot d$, choose values for $c$ and $d$ so that both expressions are negative.

Explanation:

Step1: Isolate $a$

$-11.2 - a = 0 \implies -a = 11.2 \implies a = -11.2$

Step2: Isolate $b$

$-1.1b = -5.5 \implies b = \frac{-5.5}{-1.1} = 5$

Step3: Isolate $c$

$2c = -7 \implies c = \frac{-7}{2} = -3.5$

Step4: Isolate $d$

$\frac{d}{-6} = 1 \implies d = 1 \times (-6) = -6$

Step5: Isolate $f$

$8.3 + f = 5.2 \implies f = 5.2 - 8.3 = -3.1$

Step6: Isolate $g$

$g + (-2) = -6.5 \implies g - 2 = -6.5 \implies g = -6.5 + 2 = -4.5$

Step7: Choose $e,d$ for negative expressions

We need $e-d < 0$ (so $e < d$) and $e \cdot d < 0$ (so one positive, one negative). Let $e=-2$, $d=3$:
$e-d = -2-3=-5 < 0$, $e \cdot d = -2 \times 3=-6 < 0$

Answer:

  1. $a = -11.2$
  2. $b = 5$
  3. $c = -3.5$
  4. $d = -6$
  5. $f = -3.1$
  6. $g = -4.5$

For the expressions $e-d$ and $e \cdot d$: Example values are $e=-2$ and $d=3$ (other valid pairs exist, e.g., $e=1, d=-2$ does not work; valid pairs require $e