a) To complete the table of values for the equation 2x + y + 1 = 0, you can substitute the given x-values into the equation to find the corresponding y-values. Here's the completed table of values:
x | y
---------
-4 | 9
-1 | 4
1 | -3
2 | -5
b) To draw the graph of the equation 2x + y + 1 = 0, we can use the completed table of values. On the graph paper provided, with a scale of 1 cm to 1 unit, plot the points (-4, 9), (-1, 4), (1, -3), and (2, -5). Then, draw a straight line connecting these points.
c) For the line that goes through the point (6, 7) and has a gradient of 3, we can use the slope-intercept form of a linear equation, y = mx + b, where m represents the gradient and b represents the y-intercept.
Since the gradient is given as 3, the equation becomes y = 3x + b.
To find the value of b, we can substitute the coordinates of the given point into the equation:
7 = 3(6) + b.
Solving this equation, we find that b = -11.
Therefore, the equation of the line is y = 3x - 11.
d) To express the equation y = 3x - 11 in the form Ax + By = C, we can rearrange the equation:
3x - y = 11.
In this form, A = 3, B = -1, and C = 11.
Therefore, the equation of this line in the required form is 3x - y = 11.
e) To find the coordinates of the point where the two lines intersect, we can solve the system of equations formed by equating the two equations.
The system of equations is:
2x + y + 1 = 0,
3x - y = 11.
By solving this system of equations, we can find the values of x and y at the point of intersection.