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Question: How do you find the x and y intercept of \[3x + 4y = 12\] ?...

How do you find the x and y intercept of 3x+4y=123x + 4y = 12 ?

Explanation

Solution

Hint : 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 answer :
Given,
3x+4y=123x + 4y = 12 .
To find the x-intercept we substitute y=0y = 0 in the given equation we have,
3x+4(0)=123x + 4(0) = 12
3x=123x = 12
Dividing by 3 on both side of the equation we have
x=123x = \dfrac{{12}}{3}
x=4x = 4
That is x-intercept is 4.
To find the y-intercept we substitute x=0x = 0 in the given equation we have,
3(0)+4y=123(0) + 4y = 12
4y=124y = 12
Divide the whole equation by 4 we have,
y=124y = \dfrac{{12}}{4}
y=3y = 3 .
That is y-intercept is 3.
Thus, we have the x-intercept is 4. The y-intercept is 3.
So, the correct answer is “ x-intercept is 4 and The y-intercept is 3”.

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,
3x+4y=123x + 4y = 12
We need 1 on the right hand side of the equation. So we divide the equation by 12 on both sides.
3x+4y12=1212\dfrac{{3x + 4y}}{{12}} = \dfrac{{12}}{{12}}
Separating the terms in the left hand side of the equation. We have,
3x12+4y12=1212\dfrac{{3x}}{{12}} + \dfrac{{4y}}{{12}} = \dfrac{{12}}{{12}}
Now cancelling we have,
x4+y3=1\dfrac{x}{4} + \dfrac{y}{3} = 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.