No Solution, Infinite Solutions
Date: 01/03/2002 at 21:03:08 From: Corinne Subject: Two algebra problems (1) 8(2x - 3) = 4(4x - 8) 16x - 24 = 16x - 32 16x - 16x - 24 = 16x - 16x - 32 -24 = -32 The 16x cancels so that there is no x left. How can this be solved? (2) -3(x - 3) >= 5-3x -3x + 9 >= 5-3x -3x + 3x + 9 >= 5-3x+3x 9 >= 5 The x's cancel and this seems unsolvable. Can it be solved and put on a number line?
Date: 01/03/2002 at 22:50:57 From: Doctor Ian Subject: Re: Two algebra problems Hi Corinne, Your first problem has no solution. One way to see that this must be true is that y = 16x - 24 is the equation of a line with slope 16 that crosses the y-axis at (0,-24), and y = 16x - 32 is the equation of a line the same slope that crosses the y-axis at (0,-32). Now, since these two lines have the same slope, they must either be the same line (in which case they intersect everywhere), or they must be parallel lines (in which case they intersect nowhere). Since they pass through the y-axis at different points, they must be different lines. So they're parallel. So they never intersect. So the equation has no solution. Does this make sense? For your second question, once again, try looking at each side of the inequality as the equation of a line. As before, the lines have the same slope, but they pass through the y-axis at two different locations, so they must be parallel. Now, in fact, the line y = -3x +9 is above the line y = -3x + 5 So in fact, this inequality is true _everywhere_... that is, for _all_ values of x. So whereas the first problem had no solution, this problem has an infinite number of solutions. Just to check that, let's pick some values for x and try them: x = 1 -3(1) + 9 >= -3(1) + 5 True x = 5 -3(5) + 9 >= -3(5) + 5 True In fact, if you try _any_ value of x, you'll get a true inequality. These two problems make a point: when you have parallel lines, you have to do a little reasoning, instead of just jumping in and trying to apply the rules of algebra. I hope this helps. Write back if you'd like to talk more about this, or anything else. - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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