If 2x + 5y =8 and 3x = 2y, what is the value of 2x + y?

Hi All,

We're told that 2X + 5Y = 8 and 3X = 2Y. We're asked for the value of 2X + Y. This prompt gives us two variables and two unique equations, so it's a "System" question. There are a variety of different ways that you can approach the Algebra - and based on the answer choices, the math won't involve nice, "round" numbers (in that way that most System questions do).

I'll approach this using Substitution. Since 3X = 2Y, we know that 1.5X = 1Y. We can plug that into the first equation, which gives us...

2X + 5(1.5X) = 8
2X + 7.5X = 8
9.5X = 8
19X = 16
X = 16/19

While this is not 'pretty', it does match up with how most of the answers are written (four of the answers involve "nineteenths"). With this value of X, we can solve for Y...

3X = 2Y
3(16/19) = 2Y
48/19 = 2Y
24/19 = Y

With the value of X and the value of Y, we can answer the question that's asked:
2X + Y = ?
2(16/19) + 24/19 =
32/19 + 24/19 =
56/19

Final Answer:

GMAT assassins aren't born, they're made,
Rich

Let 2x + 5y = 8 be Eq. [1]. Rearranging the terms of the second equation, we have:

3x - 2y = 0 → Eq. [2]

Multiplying Eq. [1] by 2, we have 4x + 10y = 16. Multiplying Eq. [2] by 5, we have 15x - 10y = 0. Adding the two new equations, we obtain:

19x = 16

x = 16/19

Substituting 16/19 for x in 3x = 2y, we have:

3(16/19) = 2y

3(8/19) = y

24/19 = y

Therefore, 2x + y = 32/19 + 24/19 = 56/19.

Answer: D­

\(2x + 5y = 8\)
\(4x + 10y = 16\)

\(15x = 10y\)

\(15x + 4x = 16\)
\(19x = 16\)
\(x = \frac{16}{19}\)

\(2(\frac{16}{19}) + 5y = 8\)
\(\frac{32}{19} + 5y = 8\)
\(5y = \frac{120}{19}\)
\(y = \frac{24}{19}\)

\(2x + y = \frac{32}{19} + \frac{24}{19} = \frac{56}{19}\)­

Eliminate all the x present in the equations and in the question 2x + y
Do everything since the beginning in terms of y.
Other questions can be solved with the same method.
GMAT is about technique, but you need method and didactic to know the correct technique.

I'm a fan of avoiding most of the fractions by multiplying equations to get a common coefficient for one of the variables.
Multiply 2x + 5y = 8 by 2 --> 4x + 10y = 16
Multiply 3x = 2y by 5 --> 15x = 10y
Substitute 15x for the 10y in the first equation --> 4x + 15x = 16 --> 19x = 16 --> x = 16/19
3(16/19) = 2y --> y = 24/19
2x+y = 56/19

Answer choice D.­

Since 3x = 2y,
Thus x = 2/3y...(i)

Also given, 2x + 5y = 8
Putting value of (i) in the equation,
4/3y + 5y = 8
Or, 19y = 24, y = 24/19...(ii)

And since x = 2/3y, 2x = 96/57y , y = 32/19....(iii)

From equation (ii) and (iii),
2x + y = 32/19 + 24/19
= 56/19

Thus, the correct option is D.

y = 3x/2
2x + 5*3x/2 = 8
x = 16/19
y = 24/19
2x+y = 32/19 + 24/19 = 56/19 hence D.

I wonder if there is a non-calculation intensive solution to manage this, or at least minimize the calculations, as the avg. time on this q is quite high, for the difficulty level.

KarishmaB Bunuel

The question is testing whether you know how to solve a system of simultaneous equations. The use of elimination method is better here since one of the equations doesn't have a constant term.

I need the value of 2x + y so I will keep that in mind.

Start with the second equation 3x = 2y
Multiply by 2 and divide by 3 to get
\(2x = \frac{4y}{3}\) (Equation 1 has 2x and the expression whose value we need also has 2x )

Replace 2x in equation (1) to get \(\frac{4y}{3} + 5y = 8\) so y = 24/19

We have the value of y now so let's get the value of 2x.
\(2x = (\frac{4}{3}) * (\frac{24}{19}) = \frac{32}{19}\)

\(2x + y = \frac{32}{19} + \frac{24}{19} = \frac{56}{19}\)

Answer (D)

Here is a video on how to solve linear simultaneous equations: https://youtu.be/Nh77CobN9mQ

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