
How do you graph $57-4t\ge B?$
Answer
544.8k+ views
Hint: Graphing an inequality on a number line is very similar to graphing a number.
The inequalities can be solved by algebraic operation. For simplifying the inequality the algebraic operation required. Then it is simply plotted like numbers on a number line.
Complete step-by-step answer:
The given inequality is:
$57-4y\ge 13...(i)$
$'\ge '$ This sign indicates that the left hand side value is greater than or equal to the right hand side value. The $57\text{ }8\text{ }13$ is a constant term and $4t$ is the variable term.
In this question we discuss. How to graph.
$57-4t\ge 13...(ii)$
So,
We have to simplify the given term by some algebraic operation.
Hence,
Add $4t$ and subtract $13$ from both sides of the inequality so we get,
$57-4t\ge 13$
$57-4t+4t-13\ge 13-13+4t$
From this algebraic operation.
We get,
$44\ge 4t.$
By simplifying this equation.
Hence we get.
$4t\le 11$
The simplest way to graph this is on a number line
$57-4x\ge 13$
Hence in this way graphs can be drawn.
Additional Information:
There are three methods for solving inequalities.
Explanation:
(1) First method:
By algebraic method
Example: solve: $2x-7$2x-x<7-5$
$x<2$
(2) Second method:
By number line method:
Example solve $f(x)={{x}^{2}}+2x-3<0$
First, solve $f(x)=0$ There are $2$real roots ${{x}_{1}}=1\And {{x}_{2}}=-3$
Replace $x=0$ into $f(x)$ we find $E(0)=-3<0$
Therefore, the origin $'0'$ is located inside the solution set
Answer by interval: $\left( -3,1 \right)$
(3) By graphing method:
Example: Solve $f(x)={{x}^{2}}+2x-3<0$
The graph of $f(x)$ is an upward parabola $\left( a>0 \right),$ that intersects the $x$-axis at ${{x}_{1}}=1$ and ${{x}_{2}}=-3$. Inside the interval $\left( -3,1 \right)$ the parabola stay below the $ay>>f(x)<0$
Therefore, the solution set is the open interval $\left( -3,1 \right)$
Graph $\left\{ {{x}^{2}}+2x-3\left[ -10,10,-5,5 \right] \right\}$
Note:
Always make sure while writing numbers on a number line makes comparing numbers easier. Numbers on the left are smaller than the numbers on the right of the number line.
While solving inequality, during algebraic operation add or subtract the same number to both sides, or multiply or divide the same number to both sides.
First check the type of equation term. If the $\ne ,>,\ge ,<,\le $ are signed in two sides then that must be an inequality equation term.
$\ne $ not equal to
$>$ greater than
$\ge $ greater than or equal to
$<$ less than
$\le $ less than or equal to
The inequalities can be solved by algebraic operation. For simplifying the inequality the algebraic operation required. Then it is simply plotted like numbers on a number line.
Complete step-by-step answer:
The given inequality is:
$57-4y\ge 13...(i)$
$'\ge '$ This sign indicates that the left hand side value is greater than or equal to the right hand side value. The $57\text{ }8\text{ }13$ is a constant term and $4t$ is the variable term.
In this question we discuss. How to graph.
$57-4t\ge 13...(ii)$
So,
We have to simplify the given term by some algebraic operation.
Hence,
Add $4t$ and subtract $13$ from both sides of the inequality so we get,
$57-4t\ge 13$
$57-4t+4t-13\ge 13-13+4t$
From this algebraic operation.
We get,
$44\ge 4t.$
By simplifying this equation.
Hence we get.
$4t\le 11$
The simplest way to graph this is on a number line
$57-4x\ge 13$
Hence in this way graphs can be drawn.
Additional Information:
There are three methods for solving inequalities.
Explanation:
(1) First method:
By algebraic method
Example: solve: $2x-7
$x<2$
(2) Second method:
By number line method:
Example solve $f(x)={{x}^{2}}+2x-3<0$
First, solve $f(x)=0$ There are $2$real roots ${{x}_{1}}=1\And {{x}_{2}}=-3$
Replace $x=0$ into $f(x)$ we find $E(0)=-3<0$
Therefore, the origin $'0'$ is located inside the solution set
Answer by interval: $\left( -3,1 \right)$
(3) By graphing method:
Example: Solve $f(x)={{x}^{2}}+2x-3<0$
The graph of $f(x)$ is an upward parabola $\left( a>0 \right),$ that intersects the $x$-axis at ${{x}_{1}}=1$ and ${{x}_{2}}=-3$. Inside the interval $\left( -3,1 \right)$ the parabola stay below the $ay>>f(x)<0$
Therefore, the solution set is the open interval $\left( -3,1 \right)$
Graph $\left\{ {{x}^{2}}+2x-3\left[ -10,10,-5,5 \right] \right\}$
Note:
Always make sure while writing numbers on a number line makes comparing numbers easier. Numbers on the left are smaller than the numbers on the right of the number line.
While solving inequality, during algebraic operation add or subtract the same number to both sides, or multiply or divide the same number to both sides.
First check the type of equation term. If the $\ne ,>,\ge ,<,\le $ are signed in two sides then that must be an inequality equation term.
$\ne $ not equal to
$>$ greater than
$\ge $ greater than or equal to
$<$ less than
$\le $ less than or equal to
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