course Mth 158 `syFŞҒEfxassignment #010
......!!!!!!!!...................................
16:22:17 query 1.1.20 (was 1.1.12). Explain, step by step, how you solved the equation 5y + 6 = -18 - y
......!!!!!!!!...................................
RESPONSE --> subtract 6 from both sides 5y+6 (-6)=-18 - y (-6) 5y=12-y add y to both 5y +y = 12-y +y 6y=12 divide both sides by 6 y=2
.................................................
......!!!!!!!!...................................
16:22:22 ** STUDENT SOLUTION WITH INSTRUCTOR COMMENT: 5y + 6 = 18 - y Subtract 6 from both sides, giving us 5y = 12 - y Add y to both sides, 5y + y = 12 or 6y = 12 divide both sides by 6 y = 2 INSTRUCTOR COMMENT: This is correct for equation 5y + 6 = 18 - y but the equation as I note it is 5y + 6 = -18 - y. If that's the correct equation the solution is found by practically the same steps but you end up with y = -4. **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:24:48 query 1.1.38 \ 44 (was 1.1.30). Explain, step by step, how you solved the equation (2x+1) / 3 + 16 = 3x
......!!!!!!!!...................................
RESPONSE --> multiply both by 3 2x+1 + 48=9x subtract 2x from both 1+48 = 7x 49=7x dvide both sides by 7 x=7
.................................................
......!!!!!!!!...................................
16:24:50 ** STUDENT SOLUTION: (2x + 1) / 3 + 16 = 3x First, multiply both sides of the equation by 3 2x +1 + 48 =9x or 2x + 49 = 9x subtract 2x from both sides. 49 = 7x Divide both sides by 7 7 = x **
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
16:29:41 query was 1.1.44 \ 36. Explain, step by step, how you solved the equation (x+2)(x-3) = (x+3)^2
......!!!!!!!!...................................
RESPONSE --> x^2 - x -6 = x^2 + 6x + 9 take away x^2 form both -x - 6 = 6x +9 subtract 9 from both sides -x-6-9=6x+9-9 -x-15=6x add x to both 15=7x -15/7=x
.................................................
......!!!!!!!!...................................
16:29:46 ** STUDENT SOLUTION: (x+2)(x+3) = (x+3)^2 First, we use the distributive property to remove the parenthesis and get x^2 - x - 6 = x^2 + 6x + 9 subtract x^2 from both sides, -x - 6 = 6x + 9 Subtract 9 from both sides - x - 6 - 9 = 6x or -x - 15 = 6x add x to both sides -15 = 7x Divide both sides by 7 -15/7 = x **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:31:52 query 1.1.52 (was 1.1.48). Explain, step by step, how you solved the equation x / (x^2-9) + 4 / (x+3) = 3 / (x^2-9)/
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
16:32:08 ** Starting with x / (x^2 -9) + 4 / (x+3) = 3 / (x^2 -9), first factor x^2 - 9 to get x / ( (x-3)(x+3) ) + 4 / (x+3) = 3 / ( (x-3)(x+3) ). Multiply both sides by the common denominator ( (x-3)(x+3) ): ( (x-3)(x+3) ) * x / ( (x-3)(x+3) ) + ( (x-3)(x+3) ) * 4 / (x+3) = ( (x-3)(x+3) ) * 3 / ( (x-3)(x+3) ). Simplify: x + 4(x-3) = 3. Simplify x + 4x - 12 = 3 5x = 15 x = 3. If there is a solution to the original equation it is x = 3. However x = 3 results in denominator 0 when substituted into the original equation, and division by 0 is undefined. So there is no solution to the equation. When you multiplied both sides by x-3, if x = 3 you were multiplying by zero, which invalidated your solution **
......!!!!!!!!...................................
RESPONSE --> ok
.................................................
......!!!!!!!!...................................
16:36:48 ** STUDENT SOLUTION: 1) clear the equation of fractions by multiplying both sides by the LCM (10w - 7)(5W + 7) After cancellation the left side reads: (5w+7)(8w + 5) After cancellation the right side reads: (10w - 7)(4w - 3) multiply the factors on each side using the DISTRIBUTIVE LAW Left side becomes: (40w^2) + 81w + 35 Right side becomes: (40w^2) - 58w + 21 3) subtract 40w^2 from both sides add 58w to both sides subtract 35 from both sides Rewrite: 139w = - 14 Now divide both sides by 139 to get w = - (14 / 139) DER**
......!!!!!!!!...................................
RESPONSE --> (5w+7)(8w+5) = (10w-7)(4w-3) work what you can 40w^2 + 35 = 40w^2 +21 take away 40w^2 from both sides didnt understand this one..;
.................................................
......!!!!!!!!...................................
16:36:55 query 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
......!!!!!!!!...................................
RESPONSE -->
.................................................
......!!!!!!!!...................................
16:37:07 query 1.1.70 (was 1.1.78). Explain, step by step, how you solved the equation 1 - a x = b, a <> 0.
......!!!!!!!!...................................
RESPONSE --> not sure why my answer and question cleared here...
.................................................
......!!!!!!!!...................................
16:37:10 ** Start with 1 -ax = b, a <> 0. Adding -1 to both sides we get 1 - ax - 1 = b - 1, which we simplify to get -ax = b - 1. Divide both sides by -a, which gives you x = (b - 1) / (-a). Multiply the right-hand side by -1 / -1 to get x = (-b + 1) / a or x = (1 - b) / a. **
......!!!!!!!!...................................
RESPONSE -->
.................................................
"