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{:[3y >= 18 x+6],[20 x+y >= 210]:}
If
(8,b) is a solution to the system of inequalities, what is the minimum value of
b ?

$\begin{aligned} 3 y & \geq 18 x+6 \\ 20 x+y & \geq 210 \end{aligned}$ $\newline$ If $(8, b)$ is a solution to the system of inequalities, what is the minimum value of $b$ ?

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Is (x, y) a solution to the system of linear inequalities?

Full solution

Q.

\begin{aligned} 3 y & \geq 18 x+6 \\ 20 x+y & \geq 210 \end{aligned}

\newline

If

(8, b)

is a solution to the system of inequalities, what is the minimum value of

b

?

Identify system and point: Identify the system of inequalities and the given point that is a solution to the system. $\newline$ The system of inequalities is: $\newline$ $1$ . $3y \geq 18x + 6$ $\newline$ $2$ . $20x + y \geq 210$ $\newline$ The given point is $(8, b)$ .
Substitute $x$ into first inequality to find the value of $b$ : Substitute $x = 8$ into the first inequality to find the corresponding value of $y$ (which is $b$ in this case). $\newline$ $3b \geq 18(8) + 6$ $\newline$ Perform the multiplication and addition to solve for b. $\newline$ $3b \geq 144 + 6$ $\newline$ $3b \geq 150$ $\newline$ Divide both sides of the inequality by $3$ to isolate $b$ . $\newline$ $b \geq \frac{150}{3}$ $\newline$ $b \geq 50$
Substitute $x$ into second inequality to find the value of $b$ : Now substitute $x = 8$ into the second inequality to check if it gives a higher minimum value for $b$ . $\newline$ $20(8) + b \geq 210$ $\newline$ $160 + b \geq 210$ $\newline$ Subtract $160$ from both sides of the inequality to solve for $b$ . $\newline$ $b \geq 210 - 160$ $\newline$ $b \geq 50$
Compare minimum values: Compare the minimum values of $b$ obtained from both inequalities. $\newline$ From the first inequality, we have $b \geq 50$ . $\newline$ From the second inequality, we also have $b \geq 50$ . $\newline$ Since both inequalities give the same minimum value for $b$ , the minimum value of $b$ that satisfies both inequalities is $50$ .

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