y = 2x-5 4x-y = 7 => y = 4x-7 2x-5 = 4x-7 2x-4x = 5-7 -2x = -2 x = 1 Therefore by substituting: x = 1 and y = -3
y = 2x -1 y = 5x -13 So: 5x -13 = 2x -1 5x -2x = 13 -1 3x = 12 x = 4 and by substituting y = 7
x + 2y = 7 2y = -x + 7 y = -(x-7)/2 => -x/2 + 7/2 2x - y = 7 -y = -2x + 7 y = 2x + 7 Since -1/2 is the negative reciporical of 2, the slopes of these equations are perpendicular. Therefore, these two lines are perpendicular.
(1, -1)
Because the slope of these lines are the same, they are parallel. One crosses the y-axis at 7 and the other at -7. When written in this manner the number in front of the x is the slope.
3
The vertex is at (-1,0).
-5
y = 2x-5 4x-y = 7 => y = 4x-7 2x-5 = 4x-7 2x-4x = 5-7 -2x = -2 x = 1 Therefore by substituting: x = 1 and y = -3
(-3, -5)
Need signs between terms.
y=2x+7. The 2 is your slope and the 7 is your y intercept.
y = 3x2+2x-1 Line of symmetry: x = -1/3 Vertex coordinate: (-1/3, -4/3)
2x = y + 7 x - 2 = y substitute the y from the second equation into the first: 2x = x - 2 + 7 2x = x+5 x = 5 --> y = 5 - 2 y = 3
y=f(x)= x(2x+y)=7 f(x)=2x^2 +xy -7 = 0 y=2x^2 +xy -7 y-xy -7 + 2x^2 = 0 y(1-x)=2x^2-7 y=(2x^2-7) / (1-x) Excuse the working. Y equals 2 X squared minus 7 all divided by 1-X
12
y = 2x -1 y = 5x -13 So: 5x -13 = 2x -1 5x -2x = 13 -1 3x = 12 x = 4 and by substituting y = 7