Question

question 2
translate the verbal inequality to an algebraic inequality. then solve it.
the sum of twice a number and 10 is less than three times the number increased by 4.
$2x + 10 \leq 3x + 4; x \geq 6$
$2x - 10 \geq 3x + 4; x \leq -14$
$2x + 10 < 3x + 4; x > 6$
$2x - 10 < 3x + 4; x > -14$
question 3
translate and solve the inequality.
a number increased by six is less than half that number.
(fill each blank with an integer or inequality symbol.)
x \square \square

Explanation:

QUESTION 2

Step1: Define variable

Let $x$ = the unknown number.

Step2: Translate verbal to algebraic

Sum of $2x$ and 10: $2x+10$. Less than $3x+4$: $2x+10 < 3x+4$.

Step3: Solve for $x$

Subtract $2x$ from both sides: $10 < x+4$. Subtract 4: $x > 6$.

Step1: Define variable

Let $x$ = the unknown number.

Step2: Translate verbal to algebraic

$x+6 < \frac{1}{2}x$

Step3: Solve for $x$

Subtract $\frac{1}{2}x$: $\frac{1}{2}x +6 < 0$. Subtract 6: $\frac{1}{2}x < -6$. Multiply by 2: $x < -12$.

Answer:

$2x + 10 < 3x + 4; x > 6$

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QUESTION 3