anonymous
  • anonymous
Can Someone Help Me Understand If x/5-4/3=2 a Linear Equation or not
Mathematics
  • Stacey Warren - Expert brainly.com
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chestercat
  • chestercat
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anonymous
  • anonymous
A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and (the first power of) a single variable. In our case: \[\frac{ x }{ 5 }-\frac{ 4 }{ 3 }=2 \]This equation only has *one* variable and it's in *first* power. It also has a single constant. Note: If drawn on a Cartesian graph, this line would be perfectly vertical at a specified x-value. Hence, it's already a line.
anonymous
  • anonymous
If you want to identify linear equations, the main thing to look at would be the equation's variable. For linear equations, you want only one variable in its first power (in most cases, it's x).
anonymous
  • anonymous
sorry not following @Cardinal_Carlo

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anonymous
  • anonymous
Okay, let's make this simple. Look at the only variable. \[x = x ^{1}\]One variable (x), and it's in the first power (1). Therefore, linear equation.
anonymous
  • anonymous
so that makes the equation a linear equation
anonymous
  • anonymous
Precisely. @Eianna1307 Also note: \[x ~\rightarrow~ linear\]\[x ^{2} ~\rightarrow~ quadratic\]\[x ^{3} ~\rightarrow~ cubed\]
anonymous
  • anonymous
I Thought \[x ^{2} = cubed \]\[x ^{3}= quadratic\]
anonymous
  • anonymous
^_^ no that's backwards.
anonymous
  • anonymous
A cube has 3 dimensions. A square (or quadrant) only has 2 dimensions. Contrarily, a line only has 1 dimension.
anonymous
  • anonymous
So If Its Linear How Would You Put It In Standard Form Without The Y @Cardinal_Carlo
anonymous
  • anonymous
So, standard form is this: \[Ax + By = C\]And since, y is nonexistent, we can just let B = 0. Like this: \[\frac{ 1 }{ 5 }x + 0y = \frac{ 10 }{ 3 }\]
anonymous
  • anonymous
Got Confused When You Got \[\frac{ 10 }{ 3 }\] out of \[\frac{ 4 }{ 3 }\]
anonymous
  • anonymous
@Cardinal_Carlo
anonymous
  • anonymous
Okay, look: \[\frac{ x }{ 5 }-\frac{ 4 }{ 3 }=2\]\[\frac{ x }{ 5 }=2 + \frac{ 4 }{ 3 } = \frac{ 2(3) + 4 }{ 3 } = \frac{ 6 + 4 }{ 3 } = \frac{ 10 }{ 3 }\]\[\frac{ x }{ 5 } = \frac{ 10 }{ 3 }\]\[\frac{ 1 }{ 5 }x = \frac{ 10 }{ 3 }\] And then, we add a placeholder for y to get a standard form: \[\frac{ 1 }{ 5 } x + 0y = \frac{ 10 }{ 3 }\]
anonymous
  • anonymous
That Makes Sure More Thank you @Cardinal_Carlo
anonymous
  • anonymous
You're welcome

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