15 less than y the difference of 15 and y
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9>y≥5
Absolute value of (y - x) is equal to or less than 9 . |y-x| ≤ 9
Let the two numbers be ( x ) and ( y ), where ( x - y = 9 ) and ( x + y = 137 ). By solving these equations simultaneously, we can express ( x ) as ( y + 9 ). Substituting this into the second equation gives ( (y + 9) + y = 137 ), leading to ( 2y + 9 = 137 ). Solving for ( y ) results in ( y = 64 ) and consequently ( x = 73 ). Thus, the two numbers are 73 and 64.
15 less than y the difference of 15 and y
y + 9 x + 9 ( note that "y" and "x" are both variables in algebra, unless you are working with graphs)
9 < y
x=9; y=4; x=x+y; #x=13;y=4; y=x-y; #y=13-4=9; x=x-y; #x=13-9=4;
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You need two independent linear equations to solve for two variables, x and y. The most you can do with just one equation is to simplify: y + 5x = 9 and then write one variable in terms of the other, if that is of any use. y = 9 - 5x or x = (9 - y)/5
9>y≥5
Absolute value of (y - x) is equal to or less than 9 . |y-x| ≤ 9
-3x+9=y
Let the two numbers be ( x ) and ( y ), where ( x - y = 9 ) and ( x + y = 137 ). By solving these equations simultaneously, we can express ( x ) as ( y + 9 ). Substituting this into the second equation gives ( (y + 9) + y = 137 ), leading to ( 2y + 9 = 137 ). Solving for ( y ) results in ( y = 64 ) and consequently ( x = 73 ). Thus, the two numbers are 73 and 64.
xy = 20 x+y=9 x=9-y y(9-y) = 20 9y-y(squared) = 20 0= y(squared)-9y+20 0= (y-5)(y-4) y= 5 or 4, x= 5 or 4 The two numbers are 5 and 4.
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