Answer:
32) (x, y) = (2, 4)
34) (x, y) = (-2, 3)
35) (x, y) = (1, 2)
Explanation:
32) It can be helpful to remember that "intercept form" is ...
... x/(x-intercept) +y/(y-intercept) = 1
That is, if you divide each equation by the constant on the right and express the x- and y-coefficients as denominators, then those denominators are the x- and y-intercepts of the line. Knowing those can simplify graphing.
The first equation can be rewritten as ...
... x/(-2) +y/2 = 1 . . . . . . intercepts are (-2, 0) and (0, 2)
The second equation can be rewritten as ...
... x/6 +y/6 = 1 . . . . . . . intercepts are (6, 0) and (0, 6)
Then the two lines can be graphed as in the attachment, and the solution (point of intersection) found to be (x, y) = (2, 4).
34) When you use "substitution", you use an expression for a variable in place of that variable. Here, y already has an expression written for it:
... y = x + 5
Substitute that expression for y in the other equation:
... 3x + (x+5) = -3 . . . . . . . substitute x+5 for y
... 4x = -8 . . . . . . . . . . . . . simplify, subtract 5
... x = -2 . . . . . . . . . . . . . . divide by the x-coefficient
... y = x+5 = -2+5 = 3
The solution is (x, y) = (-2, 3).
35) The y-coefficients match, so it is convenient to subtract one equation from the other to eliminate the y-variable. Here, the second equation has the smallest x-coefficient, so it will work well to subtract the second equation.
... (3x -2y) -(-2x -2y) = (-1) -(-6)
... 5x = 5 . . . . simplify. The y-variable has been eliminated.
... x = 1 . . . . . . divide by the coefficient of x
... 3(1) -2y = -1 . . . . substitute for x in the first equation
... -2y = -4 . . . . . . . subtract 3
... y = 2 . . . . . . . . . divide by the coefficient of y
The solution is (x, y) = (1, 2).