 2021-12-20

Using algebra, find the solution to the system of inequalities below
$y\le 5-3x$
$y+19\ge {x}^{2}-8x$ Jonathan Burroughs

Step 1
Given, $y\le 5-3x,y+19\ge {x}^{2}-8x$
Step 2
$y+19-19\ge {x}^{2}-8x-19$ [Subtract by 19]
$y\ge {x}^{2}-8x-19$
${x}^{2}-8x-19\le y$
So, ${x}^{2}-8x-19\le y\le 5-3x$
${x}^{2}-8x-19-5+3x\le 5-3x-5+3x$ [Subtract 5 and add 3x]
${x}^{2}-8x+3x-24\le 0$
$x\left(x-8\right)+3\left(x-8\right)\le 0$
$\left(x+3\right)\left(x-8\right)\le 0$
Step 3
The region is possible when
So, $-3\le x\le 8$. In interval notation, $x\in \left[-3,8\right]$
Step 4
Now, $\left(-3\right)\left(-3\right)\ge -3x\ge \left(-3\right)8$ [Multiply by -3]
$9\ge -3x\ge -24$
$-24\le -3x\le 9$
$5-24\le 5-3x\le 5+9$ [Add 5]
$-19\le y\le 14$
In interval notation, $y\in \left[-19,14\right]$
Step 5
Hence, the solution of system of inequalities is $x\in \left[-3,8\right],y\in \left[-19,14\right]$. Becky Harrison

Step 1
Given: $y\le 5-3x$ (1)
$y+19\ge x-8x$ (2)
Step 2
Simplify the equation (2)
$y+19\ge x-8x$
$y+19\ge -7x$
$y\ge -7x-19$
$y\ge -\left(7x+19\right)$
$-y\le 7x+19$ (3)
Step 3
$0\le 4x+24$
$-24\le 4x$
$4x\ge -24$
$x\ge -6$
Step 4
Put the value of x in equation (3)
$-y\le 7\left(-6\right)+19$
$-y\le -23$
$y\ge 23$
Step 5
Answer: $\left\{\left(x,y\right)\in {R}^{2}\mid x\ge -6,y\ge 23\right\}$ nick1337

Step 1
Given the system of inequalities
$y\le 5-3x$
$y+19\ge {x}^{2}-8x$
Simplify the inequality $y+19\ge {x}^{2}-8x$ as,
$y+19\ge {x}^{2}-8x$
$y\ge {x}^{2}-8x-19$
${x}^{2}-8x-19\le 5-3x$
${x}^{2}-8x-19+3x\le 5-3x+3x$
${x}^{2}-5x-19\le 5$
${x}^{2}-5x-24\le 0$
$\left(x+3\right)\left(x-8\right)\le 0$
$-3\le x\le 8$
Step 2
Simplify the inequality $y\le 5-3x$ as,
$y\le 5-3x$
$5-3x\ge y$
$-3x\ge y-5$
$x\le \frac{-y+5}{3}$
Thus, $-3\le x\le 8$
Since $-3\le x\le 8$obtain the inequality for y as,

Thus,$-19\le y\le 14$
Therefore, the solution is

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