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Question

Question: How do you find the \(x\) and \(y\) intercepts for \(x + y = 10\)?...

How do you find the xx and yy intercepts for x+y=10x + y = 10?

Explanation

Solution

The xx intercepts for a curve on the graph are the points at which the curve intersects the x-axis. At these points y=0y = 0. Similarly, yy intercepts for a curve on the graph are the points at which the curve intersects the y-axis. At these points x=0x = 0.

Complete step by step solution:
We have to find xx and yy intercepts for the equation x+y=10x + y = 10.
First we find the xx intercept.
The xx intercepts are the points at which the curve intersects the x-axis. To find the abscissa, i.e. xx coordinate, we assume y=0y = 0 and evaluate the corresponding value of xx.
x+y=10 x+0=10 x=10  x + y = 10 \\\ \Rightarrow x + 0 = 10 \\\ \Rightarrow x = 10 \\\
Thus, we get the point as (10,0)(10,{\kern 1pt} {\kern 1pt} {\kern 1pt} 0). This is the xx intercept of the graph of the given equation.
Now we find the yy intercept.
The yy intercepts are the points at which the curve intersects the y-axis. To find the ordinate, i.e. yy coordinate, we assume x=0x = 0 and evaluate the corresponding value of yy.
x+y=10 0+y=10 y=10  x + y = 10 \\\ \Rightarrow 0 + y = 10 \\\ \Rightarrow y = 10 \\\
Thus, we get the point as (0,10)(0,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10). This is the yy intercept of the graph of the given equation.
Hence, for the graph of the given equation we get the xx intercept as (10,0)(10,{\kern 1pt} {\kern 1pt} {\kern 1pt} 0) and the yy intercept as (0,10)(0,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10).
This we can also show from the graph of the equation x+y=10x + y = 10.

We can see in the above graph that the line of the equation cuts the x-axis at (10,0)(10,{\kern 1pt} {\kern 1pt} {\kern 1pt} 0) and the y-axis at (0,10)(0,{\kern 1pt} {\kern 1pt} {\kern 1pt} 10).

Note:
To find the xx intercept we put y=0y = 0 and to find yy intercept we put x=0x = 0 in the given equation. For a linear equation, we can also find the intercepts by writing the equation in the form of xa+yb=1\dfrac{x}{a} + \dfrac{y}{b} = 1, where aa will be the xx intercept and bb will be the yy intercept. For a linear equation in two variables we get at most one xx intercept and at most one yy intercept.