Both of these are straight line equations. So just find two points on the line, and connect with a straight edge.
Do the same for the second line. Connect points (2/3,0) & (0,-2).
It looks like these two lines intersect at the point (1,1) on the graph. Plug in x=1 and y=1 and check if it satisfies both equations, and yes it does.
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∙ 12y ago(3x-4)(x+5)
x3+3x2+3x+2 divided by x+2 equals x2+x+1
3x = x + 4 x = 2 3x2=6 2+4=6
It is: 3x2+6x-11 = 0
No. Y=3X2+1 is a Binomial Equation.
(a) y = -3x + 1
x2-5-4x2+3x = 0 -3x2+3x-5 = 0 or as 3x2-3x+5 = 0
3x2 + 22x + 7 = 3x2 +21x + x + 7 = 3x(x + 7) + (x + 7) = (x + 7)(3x + 1)
(3x-4)(x+5)
x3+3x2+3x+2 divided by x+2 equals x2+x+1
3x = x + 4 x = 2 3x2=6 2+4=6
It is: 3x2+6x-11 = 0
No. Y=3X2+1 is a Binomial Equation.
(3x + 2)(x + 2) so x = -2/3 or -2
3x*x = 3x2
3x2 + 2x - 8 = 0 is a quadratic equation.
3x2 + 2x = 16 ∴ 3x2 + 2x - 16 = 0 ∴ 3x2 - 6x + 8x - 16 = 0 ∴ 3x(x - 2) + 8(x - 2) = 0 ∴ (3x + 8)(x - 2) = 0 ∴ x ∈ {-8/3, 2}