Math 017 Materials With Exercises

a)281x b)0.09xy8x c)6Solution:a) Express 81 as a square of a number 9 2 81. We writeand thus A 9x.m n mn8 4b) Recall that a a , and so x x 222 2 281x 9 x (9x. Notice that the power of x insidethe parentheses multiplied by 2 must be equal to 8. Since 42 8 , x needs to be4raised to the fourth power. In the expression 2expression4x we identify A as A x .c) Write all factors that are not already written in this form, as a square of some2 6 30.09 0.3 , 22 must be equal to 6, thus we haveThe expression 0.3x3 y y y (the power of y inside the parentheses multiplied by0.09x6y2220.3 x3 2y3y ). As a result, we have22 0.3x3 . y is written in the form0.3xA .y2A , with3Example 11.6 Write the following expressions in the form Ax By , where A , B are any numbers.Identify A and B in your representation. x 8ya)b) 3x (2y x)c) x ( y x)4Solution:a) Recall that x 8y x 8y1 x 2y. Compare the obtained4 4 4 41 1expression x 2ywith Ax By to recognize that A , B 2 .44b) We simplify the expression (hoping to obtain the desired form of theexpression). 3x (2y x) 3x 2y x 2x 2y 2x ( 2)y . Notice how, in the laststep, we expressed subtraction as the addition. Comparing 2x ( 2)y to Ax Bygives us A 2 , B 2.c) We simplify the expression x ( y x) x y x y 0x ( 1)y .Notice that after the simplification the variable x has „disappeared‟ from theexpression. Since zero multiplied by any quantity is equal to zero, in such situation, wecan write 0 x to match the prescribed form. This „trick‟ is often used, and thus worthremembering. The comparison of 0x ( 1)y with Ax By gives us A 0,B 1.Example 11.7 Write the following equation in the form y mx b , where m and b are any numbers.In the expression you obtained, identify m and b .a) y 2x 4b) y x 0c) 2y 4x 3)2107