To simplify, you could put the equation into slope-intercept form (y=mx+b) giving you y=3/2x+1/2. From there you put in values of x and the equation gives you y-values. take each coordinate pair and place it on the graph:
X Y
1 2
2 3.5
3 5
4 6.5
5 8
from there, the slope would be 3/2, as in it goes up 3 units every time it goes over 2 units, and it intersects the y-axis at 1/2.
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
A straight line through the points (0, 1) and (-0.4, 0).
The elimination method only works with simultaneous equations, hence another equation is needed here for it to be solvable.
The equation "x plus 2y equals 5" is: x + 2y = 5 2y = 5 - x y = 5/2 - 1/2x y = -1/2x + 5/2
Get in function form. - 3X - Y = 4 - 1(- Y = 3X + 4) Y = - 3X - 4 ----------------------------solve for X and Y by the 0 out method - 3X - 4 = 0 - 3X = 4 X = - 4/3 --------------- Y = - 3(0) - 4 Y = - 4 -------------- Draw a line linking those points.
The lines are parallel.
3x+2y = 1 2y = -3x+1 y = -1.5x+0.5
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
3x = 2y + 3 3x -3 = 2y +3 -3 3x - 3 = 2y (3x -3)/2 = 2y / 2 y = 1/2 x (3x - 3) 3x = 2y +3 3x - 3x = -3x +2y +3 0 = -3x +2y +3 -1 x (0) = -1 x (-3x +2y +3) 0 = 3x - 2y -3
2y + 3x = 0 2y + -3 = 0 2y = 0 + 3 , 2y = 3 y = 3/2 y = 1.5 x = -1
y=3x+(3/2), so the slope is 3 or (3/1).
(2, -1)
3x+2y=20 5x-2y=1 (3x+2y) + (5x-2y)= 20+1 3x+5x= 21....... 8x=21.........x=2 5/8 Subsitute x=3(2 5/8)+ 2y=20........... 7 7/8 +2y=20.........2y= 20- 7 7/8....... y=6 1/16
3x - 2y = -10Subtract 3x from each side:-2y = -3x - 10Multiply each side by -1:2y = 3x + 10Divide each side by 2:y = 1.5x + 5
No, it has an infinite number of solutions. The coordinates of each and every point on the line 3x + 2y + 4 = 0 is a solution.
Any graph with the slope of -1/2
Equations: 3x-2y = 1 and 3x^2 -2y^2 +5 = 0 By combining the equations into one single quadratic equation and solving it the points of contact are made at (3, 4) and (-1, -2)