Cindy selects a sample of college students. She finds a relationship between the students' scores on an impulsivity scale and th
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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.
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<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).
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<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)
