X = 8 
Y = 2 Your welcomeeeeee
        
                    
             
        
        
        
Answer:
x = 35
Step-by-step explanation:
Solve for x:
5 (x + 20) = 7 x + 30
Expand out terms of the left hand side:
5 x + 100 = 7 x + 30
Subtract 7 x from both sides:
(5 x - 7 x) + 100 = (7 x - 7 x) + 30
5 x - 7 x = -2 x:
-2 x + 100 = (7 x - 7 x) + 30
7 x - 7 x = 0:
100 - 2 x = 30
Subtract 100 from both sides:
(100 - 100) - 2 x = 30 - 100
100 - 100 = 0:
-2 x = 30 - 100
30 - 100 = -70:
-2 x = -70
Divide both sides of -2 x = -70 by -2:
(-2 x)/(-2) = (-70)/(-2)
(-2)/(-2) = 1:
x = (-70)/(-2)
The gcd of -70 and -2 is -2, so (-70)/(-2) = (-2×35)/(-2×1) = (-2)/(-2)×35 = 35:
Answer: x = 35
 
        
                    
             
        
        
        
              NOT MY WORDS TAKEN FROM A SOURCE!
(x^2) <64  => (x^2) -64 < 64-64 => (x^2) - 64 < 0 64= 8^2    so    (x^2) - (8^2) < 0  To solve the inequality we first find the roots (values of x that make (x^2) - (8^2) = 0 ) Note that if we can express (x^2) - (y^2) as (x-y)* (x+y)  You can work backwards and verify this is true. so let's set (x^2) - (8^2)  equal to zero to find the roots: (x^2) - (8^2) = 0   => (x-8)*(x+8) = 0       if x-8 = 0 => x=8      and if x+8 = 0 => x=-8 So x= +/-8 are the roots of x^2) - (8^2)Now you need to pick any x values less than -8 (the smaller root) , one x value between -8 and +8 (the two roots), and one x value greater than 8 (the greater root) and see if the sign is positive or negative. 1) Let's pick -10 (which is smaller than -8). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive
2) Let's pick 0 (which is greater than -8, larger than 8). If x=0, then (x^2) - (8^2) = 0-64 = -64 <0  so it is negative3) Let's pick +10 (which is greater than 10). If x=-10, then (x^2) - (8^2) = 100-64 = 36>0  so it is positive Since we are interested in (x^2) - 64 < 0, then x should be between -8 and positive 8. So  -8<x<8 Note: If you choose any number outside this range for x, and square it it will be greater than 64 and so it is not valid.
Hope this helped!
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Answer:
Please see attached picture
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