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Question

Question: How do you find the intercepts of \[2x + y = 6\]?...

How do you find the intercepts of 2x+y=62x + y = 6?

Explanation

Solution

Here we need to find x-intercept and y-intercept. We know that x-intercept is a point on the graph where ‘y’ is zero.
Also we know that y-intercept is a point on the graph where ‘x’ is zero. In other words the value of ‘x’ at ‘y’ is equal to zero is called x-intercept. The value of ‘y’ at ‘x’ is equal to zero is called t-intercept. Using this definition we can solve the given problem.

Complete step by step solution:
Given,
2x+y=62x + y = 6.
To find the x-intercept we substitute y=0y = 0 in the given equation we have,
2x+(0)=62x + (0) = 6
2x=62x = 6
Dividing by 2 on both side of the equation we have
x=62x = \dfrac{6}{2}
x=3x = 3
That is x-intercept is 3.
To find the y-intercept we substitute x=0x = 0 in the given equation we have,
2(0)+y=62(0) + y = 6
y=6y = 6
That is y-intercept is 6.
Thus, we have the x-intercept is 3. The y-intercept is 6.

Note: We can also solve this by converting the given equation into the equation of straight line intercept form. That is xa+yb=1\dfrac{x}{a} + \dfrac{y}{b} = 1. Where ‘a’ is a x-intercept and ‘b’ is called y-intercept.
Now given,
2x+y=62x + y = 6
We need 1 on the right hand side of the equation. So we divide the equation by 6 on both sides.
2x+y6=66\dfrac{{2x + y}}{6} = \dfrac{6}{6}
Separating the terms in the left hand side of the equation. We have,
2x6+y6=66\dfrac{{2x}}{6} + \dfrac{y}{6} = \dfrac{6}{6}
Now cancelling we have,
x3+y6=1\dfrac{x}{3} + \dfrac{y}{6} = 1.
Now comparing with the standard intercept equation we have,
The x-intercept is 4. The y-intercept is 3. In both the methods we have the same answer. We can choose any one method to solve this.