Aline has a slope of -3 and passes through the point (5,-1) write an equation of the line?
the first equation can be multiplied by 2.
the x-coefficients are -2 and 4. if the first one of these is multiplied by 2, it will be the opposite of the second of these. that is 2·(-2) = -4. when -4 and 4 are added, the result is 0, so the x-variable will be eliminated.
to solve these equations by "elimination", you look for simple relationships between the coefficients that will let you combine them to get zero.
here's the rest of the solution.
2(-2x-5y) +(4x+2y) = 2(-1) +(8) . . after multiplying the first equation by 2 and adding the second equation
-4x -10y +4x +2y = -2 +8 . . . the result of eliminating parentheses
-8y = 6 . . . the result of collecting terms
y = -6/8 = -3/4 . . divide by the y-coefficient
since all of the coefficients in the second equation are even, a factor of 2 can be removed, and the equation written as
2x +y = 4
this can be solved for x, so you have
x = (4 -y)/2 = 2 -(y/2)
filling in the above value of y, we find x to be
x = 2 /4)/2
x = 2 3/8
comment on eliminating y instead of x
if you divide the second equation by 2 so that y has a coefficient of 1, then you can multiply this equation by 5 to give y a coefficient that is the opposite of the coefficient -5 in the first equation. then the elimination looks like
(-2x -5y) +5(2x +y) = (-1) +5(4)
-2x -5y +10x +5y = -1 +20 . . eliminate parentheses
8x = 19 . . simplify
x = 19/8 = 2 3/8 . . divide by the coefficient of x
we can find the area by adding the area of the rectangle to the area of the 2 triangles.
the rectangle is
7+7+7 by 9
21 by 9
a = length * width
a = 21*9
a = 189
the triangles are identical so we find the area of one of them and multiply by 2
the base is 7 and the height is 6.1
a = 1/2 b*h
but since we have 2 of them
2a = 2*1/2 b*h
2a = b*h
2a = (7* 6.1)
add the rectangles area to the area of the two triangles
189 + 42.7
use main properties of powers: