Can Someone Help Me Understand If x/5-4/3=2 a Linear Equation or not
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.
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).
sorry not following @Cardinal_Carlo
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.
so that makes the equation a linear equation
Precisely. @Eianna1307 Also note: \[x ~\rightarrow~ linear\]\[x ^{2} ~\rightarrow~ quadratic\]\[x ^{3} ~\rightarrow~ cubed\]
I Thought \[x ^{2} = cubed \]\[x ^{3}= quadratic\]
^_^ no that's backwards.
A cube has 3 dimensions. A square (or quadrant) only has 2 dimensions. Contrarily, a line only has 1 dimension.
So If Its Linear How Would You Put It In Standard Form Without The Y @Cardinal_Carlo
So, standard form is this: \[Ax + By = C\]And since, y is nonexistent, we can just let B = 0. Like this: \[\frac{ 1 }{ 5 }x + 0y = \frac{ 10 }{ 3 }\]
Got Confused When You Got \[\frac{ 10 }{ 3 }\] out of \[\frac{ 4 }{ 3 }\]
@Cardinal_Carlo
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: \[\frac{ 1 }{ 5 } x + 0y = \frac{ 10 }{ 3 }\]
That Makes Sure More Thank you @Cardinal_Carlo
You're welcome
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