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)
Answer:
never
Step-by-step explanation:
In a right triangle, every angle has an opposite side and an adjacent side. And there's always a hypotenuse.
The sine of an angle is the ratio of the length of the opposite side divided by the length of the hypotenuse.
The tangent ratio of an angle is the opposite side over adjacent side, it is not the sine ratio of a triangle.
For example, ABC is a right angled triangle having length of the sides are a,b,c.
So
