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

Answer 1

Intercept on #x#-axis is #3# and intercept on #y#-axis is also #3#.

In #4x+4y=12#, we can find intercept on #x#-axis by putting #y=0# and intercept on #y#-axis by putting #x=0#.
Hence intercept on #x#-axis is #4x+4xx0=12# or #4x=12# or #x=3#.
and intercept on #y#-axis is #4xx0+4y=12# or #4y=12# or #y=3#.
Alternatively if intercepts on #x#-axis and #y#-axis are #a# and #b# respectively, then equation of line is #x/a+y/b=1#. Hence, we can also find intercepts on #x#-axis and #y#-axis by converting the equation in this form.
Now #4x+4y=12hArr(4x)/12+(4y)/12=1# or
#x/3+y/3=1# and hence again we get that intercept on #x#-axis is #3# and intercept on #y#-axis is also #3#.
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Answer 2

To find the x-intercept, set ( y = 0 ) in the equation and solve for ( x ). For the given equation ( 4x + 4y = 12 ), when ( y = 0 ), ( 4x = 12 ), so ( x = 3 ). Thus, the x-intercept is (3, 0).

To find the y-intercept, set ( x = 0 ) in the equation and solve for ( y ). For the given equation ( 4x + 4y = 12 ), when ( x = 0 ), ( 4y = 12 ), so ( y = 3 ). Thus, the y-intercept is (0, 3).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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