How do you solve #15y - \frac { 11} { 2} = \frac { 11} { 2} y + 2#?
You can save yourself some trouble and a possible source of error by multiplying everything by
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Substitute this value into the equation and if left side equals right side then it is the solution.
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To solve the equation ( 15y - \frac{11}{2} = \frac{11}{2}y + 2 ), you can follow these steps:
Step 1: Move all terms involving ( y ) to one side of the equation and the constant terms to the other side. You can do this by adding or subtracting terms as necessary. In this case, let's subtract ( \frac{11}{2}y ) from both sides and add ( \frac{11}{2} ) to both sides:
[ 15y - \frac{11}{2} - \frac{11}{2}y = 2 + \frac{11}{2} ]
Step 2: Simplify both sides of the equation:
[ 15y - \frac{11}{2} - \frac{11}{2}y = 2 + \frac{11}{2} ] [ (15y - \frac{11}{2}y) = (2 + \frac{11}{2}) ]
Step 3: Combine like terms:
[ (\frac{15}{2}y) = (\frac{25}{2}) ]
Step 4: Now, isolate ( y ) by dividing both sides of the equation by ( \frac{15}{2} ):
[ y = \frac{\frac{25}{2}}{\frac{15}{2}} ]
Step 5: Simplify the expression:
[ y = \frac{25}{2} \times \frac{2}{15} ] [ y = \frac{25}{15} ]
Step 6: Simplify the fraction:
[ y = \frac{5}{3} ]
So, the solution to the equation is ( y = \frac{5}{3} ).
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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