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St 1) 5x + 3y = 15 -> We get 2 values x=0 y=5 or y=0 x=3 so value of x can be 0 or 5 -> NS
St 2) y = 5 – (5/3)x -> solving this we get the same eq. as st 1's (5x + 3y = 15) so again two values of x 0 or 5 -> NS
Combining St1 and ST2 -> No unique solution!

Answer E
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(1) 5x + 3y = 15

Clearly insufficient.

(2) y = 5 – (5/3)x

Solve:
y/5= 1-x/3
x/3+y/5=1
therefore
5x+3y=15
This equation is same as equation 1)
So equation 1 and equation 2 are same and insufficient. So E
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(1) 5x + 3y = 15
no info is given on y
insufficient

(2) y = 5 – (5/3)x
If this expression is multiplied by 3, it is identical to statement 1
Insufficient

Answer : E
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Bunuel
What is the value of x?

(1) 5x + 3y = 15
(2) y = 5 – (5/3)x


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1. 5x+5y=15. Insufficient
2.y=5-(5/3)x
3y=15-5x. Insuffcient

Using both statements, data is still insufficient to find value of x
Answer E
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Bunuel
What is the value of x?

(1) 5x + 3y = 15
(2) y = 5 – (5/3)x


Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION

Since (1) is a linear equation with two variables, it will allow you to solve for x only in terms of y. So that statement is insufficient by itself, eliminate answers A and D. This leaves B, C, and E.

Since (2) is also a linear equation with two variables, it will allow you to solve for x only in terms of y. So that statement is insufficient by itself, eliminate B. This leaves C and E.

You might be tempted to choose C, because (1) and (2) together give you two linear equations with the same two variables. As we’ve seen though, we must first make sure that the two equations are distinct. To see whether they are in fact the same equation in different guises, rewrite one of them—say Statement (1)—in the form that other already has.

So let’s solve for y using (1).

5x + 3y = 15

Subtract 5x from each side.

3y=15-5x

Divide each side by 3.

y = 5 – (5/3)x

So (1) and (2) are in fact the same equation.

The correct answer is E.
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