Using algebra, find the solution to the system of inequalities below y≤5−3x y+19≥x2−8x

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

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

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,
$y+19\ge x^2-8x$
$y\ge x^2-8x-19$
$x^2-8x-19\le 5-3x$
$x^2-8x-19+3x\le 5-3x+3x$
$x^2-5x-19\le 5$
$x^2-5x-24\le 0$
$(x+3)(x-8)\le 0$
$-3\le x\le 8$
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,
$\frac{-y+5}{3}\le 8and\frac{-y+5}{3}\ge -3$
$-y\le 19and-y\ge -14\Rightarrow y\le 14$
Thus,$-19\le y\le 14$
Therefore, the solution is $-3\le x\le 8and-19\le y\le 14$