Solve the equation for the indicated variable a a 3b 3c x5...

Name: solve the equation for the indicated variable a a 3b 3c x5 for x x b ax b cx d 3 for x x c a2xa 5a5 for x x d a5b a 5b b5a for a a 80288
Uploaded: 2020-07-19T21:39:24-08:00
Duration: 7 min 37 s
Channel: Christina Shute
Description: solve the equation for the indicated variable a a 3b 3c x5 for x x b ax b cx d 3 for x x c a2xa 5a5 for x x d a5b a 5b b5a for a a 80288

Solve the equation for the indicated variable: A) a - 3[b-3(c-x)]=5; for x x= B) (ax + b )/ (cx + d )= 3; for x x= C) (a^2)x+(a-5)=(a+5); for x x= D) (a+5)/b = (a-5)/b + (b+5)/a; for a a=

Transcript

00:01 For each part of this problem, we are isolating the variable to solve for x.

00:07 So first starting on part a, we want to distribute these parentheses in the multiplication that they imply.

00:16 So we have negative 2x minus 5 minus 6 plus x plus 3 equals x minus 3x minus 6.

00:33 So then this allows us to combine like terms on each side.

00:38 So we have negative x, right? and then we have negative 11 here plus 3.

00:49 So then that's negative 8 equals negative 2x minus 6.

00:58 So i'm going to add 2x to both sides.

01:00 I get x minus 8 equals negative 6.

01:05 Add 8 to both sides then.

01:07 And we have x equals, so then that's our solution for part a.

01:17 Moving on to part b, i'm going to start by subtracting three from both sides.

01:29 So i get square root x plus 1 equals square root 2x minus 1.

01:39 The reason i'm doing this is because i'm about to square both sides, and this just makes the work a little bit simpler for me.

01:45 So, squaring both sides, i get x plus 1 multiplied by square root 2x minus 1 squared.

01:56 And that's going to take us a little bit of work.

02:00 All right, so we left alone x plus 1.

02:04 Then coming over here, we can hop up here to work this out a little bit.

02:13 I like to write these out instead of trying to do it in my head.

02:16 It helps me from getting confused.

02:19 So we have 2x minus square root 2x minus square root 2x, minus square root 2x.

02:25 2x plus 1, which equals 2x minus 2 square root 2x plus 1.

02:37 So that's what we have on this side of the equation.

02:44 So then i can continue combining like terms before dealing with this square root.

02:49 So i'm going to subtract 2x from both sides, and i get negative x plus 1 equals negative 2 square root 2x plus 1, and then i can subtract 1 from both sides.

03:02 Both sides and we get negative x equals negative 2 square root 2x.

03:10 Okay, now i'm going to square both sides once again and we get x squared equals 4 multiplied by 2x.

03:25 So performing that multiplication we get x squared equals 8x.

03:32 Then we can divide by x on both sides and we get x equals 8.

03:37 And that is our solution.

03:43 And then last but not least, we have part c...