
How do you find the x and y intercepts of \[y=3x-2\]?
Answer
556.8k+ views
Hint: This type of problem is based on the concept of equation of lines. First, we have to find the intercepts of the given equation by dividing the whole equation by 2. We then have to make necessary calculations and convert the obtained equation to \[\dfrac{x}{a}+\dfrac{y}{b}=1\]. Here, a is the x-intercept and b is the y-intercept. On comparing the obtained equation with the line-intercept equation, we get the x and y intercepts of \[y=3x-2\].
Complete answer:
According to the question, we are asked to find the intercepts of the given equation.
We have been given the equation is \[y=3x-2\]. -----(1)
Let us first divide the whole equation (1) by 2.
\[\dfrac{y}{2}=\dfrac{3x-2}{2}\]
We can now use the method \[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\] to split the R.H.S.
We get,
\[\dfrac{y}{2}=\dfrac{3x}{2}-\dfrac{2}{2}\]
On further simplification, we get,
\[\Rightarrow \dfrac{y}{2}=\dfrac{3x}{2}-1\]
Let us now subtract \[\dfrac{3x}{2}\] from both the sides of the equation.
We get,
\[\Rightarrow \dfrac{y}{2}-\dfrac{3x}{2}=\dfrac{3x}{2}-\dfrac{3x}{2}-1\]
We know that terms of opposite signs cancel out.
We get,
\[\Rightarrow \dfrac{y}{2}-\dfrac{3x}{2}=-1\]
Now divide -1 on both the sides of the obtained equation.
\[\Rightarrow \dfrac{\dfrac{y}{2}-\dfrac{3x}{2}}{-1}=\dfrac{-1}{-1}\]
\[\Rightarrow -\dfrac{y}{2}+\dfrac{3x}{2}=1\]
We know that \[\dfrac{x}{\dfrac{a}{b}}=\dfrac{bx}{a}\].
On using this identity, we get,
\[\Rightarrow \dfrac{y}{-2}+\dfrac{x}{\dfrac{2}{3}}=1\]
\[\therefore \dfrac{x}{\dfrac{2}{3}}+\dfrac{y}{-2}=1\] -------(2)
Comparing equation (2) with \[\dfrac{x}{a}+\dfrac{y}{b}=1\], we get,
\[a=\dfrac{2}{3}\] and \[b=-2\].
Therefore, the x-intercept is \[\dfrac{2}{3}\] and the y-intercept is \[-2\].
Hence, the x and y intercepts of \[y=3x-2\] are \[\dfrac{2}{3}\] and \[-2\] respectively.
Note: Whenever you get this type of problem, we should try to make the necessary calculations in the given equation to get the intercept form to find the intercepts. We should avoid calculation mistakes based on sign conventions. We can also convert the equation directly by cross multiplying the equation. Then make some necessary calculations to obtain the final answer.
Complete answer:
According to the question, we are asked to find the intercepts of the given equation.
We have been given the equation is \[y=3x-2\]. -----(1)
Let us first divide the whole equation (1) by 2.
\[\dfrac{y}{2}=\dfrac{3x-2}{2}\]
We can now use the method \[\dfrac{a+b}{c}=\dfrac{a}{c}+\dfrac{b}{c}\] to split the R.H.S.
We get,
\[\dfrac{y}{2}=\dfrac{3x}{2}-\dfrac{2}{2}\]
On further simplification, we get,
\[\Rightarrow \dfrac{y}{2}=\dfrac{3x}{2}-1\]
Let us now subtract \[\dfrac{3x}{2}\] from both the sides of the equation.
We get,
\[\Rightarrow \dfrac{y}{2}-\dfrac{3x}{2}=\dfrac{3x}{2}-\dfrac{3x}{2}-1\]
We know that terms of opposite signs cancel out.
We get,
\[\Rightarrow \dfrac{y}{2}-\dfrac{3x}{2}=-1\]
Now divide -1 on both the sides of the obtained equation.
\[\Rightarrow \dfrac{\dfrac{y}{2}-\dfrac{3x}{2}}{-1}=\dfrac{-1}{-1}\]
\[\Rightarrow -\dfrac{y}{2}+\dfrac{3x}{2}=1\]
We know that \[\dfrac{x}{\dfrac{a}{b}}=\dfrac{bx}{a}\].
On using this identity, we get,
\[\Rightarrow \dfrac{y}{-2}+\dfrac{x}{\dfrac{2}{3}}=1\]
\[\therefore \dfrac{x}{\dfrac{2}{3}}+\dfrac{y}{-2}=1\] -------(2)
Comparing equation (2) with \[\dfrac{x}{a}+\dfrac{y}{b}=1\], we get,
\[a=\dfrac{2}{3}\] and \[b=-2\].
Therefore, the x-intercept is \[\dfrac{2}{3}\] and the y-intercept is \[-2\].
Hence, the x and y intercepts of \[y=3x-2\] are \[\dfrac{2}{3}\] and \[-2\] respectively.
Note: Whenever you get this type of problem, we should try to make the necessary calculations in the given equation to get the intercept form to find the intercepts. We should avoid calculation mistakes based on sign conventions. We can also convert the equation directly by cross multiplying the equation. Then make some necessary calculations to obtain the final answer.
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