There are several methods taught for solving systems of linear equations. Among them are - elimination - substitution - graphing - Cramer's rule - matrix inversion - Gaussian elimination. The last three of these have much in common with each other and with elimination. These matrix methods probably will not be taught in a beginning algebra course. Graphing is often suitable when only an approximate answer is needed. The point of the question seems to be to have you try several of these on the given equations so you can see which seem to work better for you. It really doesn't matter which you choose, as you are to make a comparison based on your experience with the method(s). (It would make more sense to choose methods you know and understand.) Problem A Method 1 (substitution) 3x-9y=3, 6x-3y=-24 Solve the first equation for x and substitute that into the second equation. 3x = 3 + 9y (add 9y to both sides) x = 1 + 3y (divide by 3) 6(1 + 3y) - 3y = -24 (substitute the expression for x in the second equation) 6 + 18y - 3y = -24 (eliminate parentheses) 15y = -30 (collect terms, subtract 6) y = -2 (divide by 15) x = 1 + 3(-2) = -5 (use the expression for x to find x from y) Solution: (x, y) = (-5, -2) in 7 steps. Problem A Method 2 (Cramer's rule) 3x-9y = 3, 6x-3y=-24 Write the determinant expressions, then evaluate x = det[3, -9; -24, -3] / det[3, -9; 6, -3] x = ((3)(-3) - (-9)(-24)) / ((3)(-3) - (-9)(6)) = (-9 - 216)/(-9 + 54) = -225/45 = -5 y = det[3, 3; 6, -24] / 45 = ((3)(-24) - (3)(6)) / 45 = (-72 - 18)/ 45 = -90/45 = -2 Solution: (x, y) = (-5, -2) in 3 steps. I found Cramer's rule to require fewer steps, but more tedious attention to detail and math with larger numbers. Problem B Method 1 (elimination) 7x-3y=20, 5x+3y=16 Add the two equations to eliminate y. (7x-3y) + (5x+3y) = (20) + (16) 12x = 36 (collect terms) x = 3 (divide by 12) 5(3) + 3y = 16 (substitute this value into the second equation) 3y = 1 (subtract 15) y = 1/3 (divide by 3) Solution: (x, y) = (3, 1/3) in 6 steps. Problem B Method 2 (substitution) 7x-3y=20, 5x+3y=16 Solve the first equation for y. Substitute into the second equation. (By eliminating y in this way, the result is a solution that has the "feel" of elimination.) -3y = 20 - 7x (subtract 7x to get y by itself) y = -20/3 + (7/3)x (divide by -3, the coefficient of y) 5x + 3(-20/3 + (7/3)x) = 16 (perform the substitution into the second equation) 12x - 20 = 16 (collect terms) x = 36/12 = 3 (add 20, divide by 12) y = -20/3 + (7/3)(3) = (-20+21)/3 = 1/3 Solution: (x, y) = (3, 1/3) in 6 steps (or so). In my opinion, the methods work out about the same, but substitution in this case involved more work with fractions. Problem C Method 1 (substitution) y = (1/2)x-6, 2x+6y=19 As there is already an expression for y, substitution is the "obvious" choice. 2x + 6((1/2)x-6) = 19 (use the expression for y in place of y in the second equation) 5x -36 = 19 (collect terms) 5x = 55 (add 36) x = 11 (divide by 5) y = (1/2)(11) - 6 = 5.5 - 6 = -1/2 Solution: (x, y) = (11, -1/2) in 5 steps. Problem C Method 2 (Cramer's rule) y=(1/2)x-6, 2x+6y=19 The first equation must be put into standard form. It is convenient to clear fractions, too, but not actually necessary. -(1/2)x + y = -6 x - 2y = 12 (multiply by -2 to clear fractions) x = det[12, -2; 19, 6] / det[1, -2; 2, 6] = ((12)(6) - (-2)(19))/((1)(6) - (-2)(2)) x = (72+38)/(6+4) = 110/10 = 11 y = det[1, 12; 2, 19] / 10 = ((1)(19) - (2)(12))/10 = (19-24)/10 = -5/10 = -1/2 Solution: (x, y) = (11, -1/2) in 5 steps. In my opinion, "substitution" is easier to work out in this case. It involves more fractions, but the math is not difficult. Each of these methods seems to take about the same number of steps, so one is not significantly more efficient than the other.
