+0  
 
0
7188
1
avatar

Solve for  , where is a real number.

 Jul 26, 2014

Best Answer 

 #1
avatar+4473 
+5

We begin by squaring both sides to get rid of the square roots: sqrt(6x - 15) = sqrt(4x + 7) --> 6x - 15 = 4x + 7

Next, we get all the x terms on one side and all the regular numbers to the other side by adding 15 to both sides and subtracting 4x from both sides: By doing this, we get 2x = 22.

Now, we are left with something easy and we can see that 22/11 = x --> x = 11.

Do a quick check to see if x = 11 satisfies the condition:

sqrt(6*(11) - 15) = sqrt(4*(11) + 7) 

sqrt(66 - 15) = sqrt(44 + 7)

sqrt(51) = sqrt(51) --> Yes.

 Jul 26, 2014
 #1
avatar+4473 
+5
Best Answer

We begin by squaring both sides to get rid of the square roots: sqrt(6x - 15) = sqrt(4x + 7) --> 6x - 15 = 4x + 7

Next, we get all the x terms on one side and all the regular numbers to the other side by adding 15 to both sides and subtracting 4x from both sides: By doing this, we get 2x = 22.

Now, we are left with something easy and we can see that 22/11 = x --> x = 11.

Do a quick check to see if x = 11 satisfies the condition:

sqrt(6*(11) - 15) = sqrt(4*(11) + 7) 

sqrt(66 - 15) = sqrt(44 + 7)

sqrt(51) = sqrt(51) --> Yes.

AzizHusain Jul 26, 2014

4 Online Users

avatar
avatar
avatar