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n200080 [17]
3 years ago
5

Find the solution to the linear system by using the x - and y -intercepts.

Mathematics
1 answer:
Elenna [48]3 years ago
4 0

The solution to these equations is (x, y) = (2, 4).

We've not heard of a method of solving a linear system using the x- and y-intercepts. A Google search on the subject turns up this question, and no other information. However, there is information you can gain from the intercepts that can help you find a solution. (Your reference material may provide a better source of information on this subject.)

The <em>intercept form</em> of a linear equation is ...

... x/(x-intercept) + y/(y-intercept) = 1

Dividing the first equation by 4, you can rearrange it to ...

... x/(-2) +y/(2) = 1 . . . . . . the x-intercept is -2, the y-intercept is +2.

Dividing the second equation by -6, you can rearrange it to ...

... x/-6 +y/3 = 1 . . . . . . . . the x-intercept is -6, the y-intercept is +3.

_____

<em>What you can do with the intercepts</em>

The intercepts can be used to <em>graph the equations</em>. Plot each of the intercepts for a given equation, then draw a line through them. (See the attachment.)

Here, both lines have their intercepts on the negative x and positive y axes. The slopes and intercepts of the lines are such that they intersect in the 1st quadrant.

We can use the intercepts to <em>find the slope of the line</em>.

The slope of each line can be found from ...

... slope = -(y-intercept)/(x-intercept)

Then the slope of the first line is ...

... m1 = -2/-2 = 1

and the slope of the second line is ...

... m2 = -3/-6 = 1/2

The difference in slopes is ...

... m1 - m2 = 1 - 1/2 = 1/2

Using the slope and intercept we can <em>find the solution</em> by substitution or elimination.

For line 1 with slope m1 and y-intercept b1, the equation of the line in slope-intercept form is

... y = m1·x + b1 = x +2

For line 2 with slope m2 and y-intercept b2, the equation of the line in slope-intercept form is

... y = m2·x +b2 = (1/2)x +3

Subtracting the second equation from the first eliminates the y-variable and gives ...

... y - y = 0 = (x +2) -(1/2x +3) = 1/2x - 1

Adding 1 and multiplying by 2 gives the solution for x:

... x = 2

Then the first equation gives the solution for y:

... y = x + 2 = 2+2 = 4

The solution is (x, y) = (2, 4).

_____

<em>Comment on </em><em>find the solution</em>

Above, we subtracted one equation from the other to get ...

... y - y = 0 = x(m1 -m2) +(b1 -b2)

We could have simply equated the values of y to get ...

... m1·x +b1 = y = m2·x +b2

Either way you do this, you find the solution for x is ...

... x = (b2 -b1)/(m1 -m2)

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