Quandaries and Queries I hope that u can help me....I am a college student taking a class in Pre Calculus.....I have homework due this Friday and it counts a BIG Percentage on my FINAL grade.....I am getting mixed up and can not figure out a few problems.....Please help me..... Method Of Subsitution Problem 1. y- 8x = -5 x(squared) + y(squared) = 25 Problem 2. y = x(squared) - 2x - 6 Y = x(squared) - 4 Problem 3. Solve by elimination 4x - y= -1 3x + 4y = -2 Problem 4. Method of Sub and Method of Elimination X - 2 Y + Z = 8 3X - 5Y + 10Z = 6 2X - 6Y - 11Z = 4 I hope that u can help me with this....Out of 50 questions, I got stuck on these 4.....Please help ASAP..... Thanks.. Scott Hi Scott, Since this is such a big homework assignment I am not going to do the problems for you but I will show you solutions to similar problems. Method of Substitution. y + 2x = 3 2x2 + y2 = 5 - 2x Since y + 2x = 3, y = 3 - 2x Thus, substituting into the second equation 2x2 + (3 - 2x)2 = 5 - 2x 2x2 + 9 -12x + 4x2 = 5 - 2x 6x2 - 10x + 4 = 0 2(3x2 -5x + 2) = 0 2(3x - 2)(x -1) = 0 Thus x = 2/3 or x = 1. Substitute these values back into he first equation to find the corresponding values of y. Method of elimination. 4x + 2y = 5 x - 3y = 2 Multiply the second equation by 4. This will make the coefficients of x equal in both equations. 4x + 2y = 5 4x - 12y = 8 Now subtract the second equation from the first. This gives 14y = -3 Thus y = -3/14 Substitute this value back into one of the two equations to find the corresponding value of x. For both of these problems you should check your answers. Penny Go to Math Central