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Difficulty: Very hardSystems of Linear Equations
Consider the following system of linear equations in xx, yy, and zz, where cc is a real constant:
32x−y+2z=5\frac{3}{2}x - y + 2z = 5
x+13y−z=2x + \frac{1}{3}y - z = 2
6x−y+z=c6x - y + z = c

If the system has at least one solution (x,y,z)(x, y, z), what is the value of 7x−23y7x - \frac{2}{3}y?

  1. 18Answer
  2. B
    16
  3. C
    24
  4. D
    7
  5. E
    2

Answer

18
To find the value of 7x−23y7x - \frac{2}{3}y without individual values for x,y,x, y, and zz, we express 7x−23y7x - \frac{2}{3}y as a linear combination m(Eq. 1)+n(Eq. 2)m(\text{Eq. 1}) + n(\text{Eq. 2}). Matching the zz-coefficients requires 2m−n=0  ⟹  n=2m2m - n = 0 \implies n = 2m. Matching the yy-coefficients yields −m+13(2m)=−23  ⟹  m=2-m + \frac{1}{3}(2m) = -\frac{2}{3} \implies m = 2, which gives n=4n = 4. Verifying the xx-coefficient gives 2(32)+4(1)=72\left(\frac{3}{2}\right) + 4(1) = 7. Applying these multipliers to the right-hand sides gives 2(5)+4(2)=10+8=182(5) + 4(2) = 10 + 8 = 18.

Step-by-Step Solution

1
Identify the target expression 7x−23y7x - \frac{2}{3}y as a linear combination of the first two equations
Express m(32x−y+2z)+n(x+13y−z)=7x−23y+0zm\left(\frac{3}{2}x - y + 2z\right) + n\left(x + \frac{1}{3}y - z\right) = 7x - \frac{2}{3}y + 0z
Because the system is dependent when consistent, individual variable values cannot be uniquely determined, but specific linear combinations independent of zz can be evaluated.
2
Set up a system of equations for the scalar multipliers mm and nn
Equating coefficients of zz: 2m−n=0  ⟹  n=2m2m - n = 0 \implies n = 2m. Equating coefficients of yy: −m+13n=−23-m + \frac{1}{3}n = -\frac{2}{3}.
Eliminating zz requires the net coefficient of zz to equal 0.
3
Solve for mm and nn
Substitute n=2mn = 2m into the yy-coefficient equation: −m+23m=−13m=−23  ⟹  m=2-m + \frac{2}{3}m = -\frac{1}{3}m = -\frac{2}{3} \implies m = 2, which gives n=4n = 4.
Determining the exact linear multipliers needed to match the target expression.
4
Verify xx-coefficient consistency and compute the target value
xx-coefficient: 2(32)+4(1)=3+4=72\left(\frac{3}{2}\right) + 4(1) = 3 + 4 = 7. Value: 2(5)+4(2)=10+8=182(5) + 4(2) = 10 + 8 = 18.
Applying the scalars m=2m = 2 and n=4n = 4 to the right-hand side constants gives the exact numerical value of 7x−23y7x - \frac{2}{3}y.

Key Concept

Linear combinations of dependent systems of equations

Alternative Method

Multiply the first equation by 2 to clear fractions: 3x−2y+4z=103x - 2y + 4z = 10. Multiply the second equation by 3 to clear fractions: 3x+y−3z=63x + y - 3z = 6. Eliminate zz by forming 3(3x−2y+4z)+4(3x+y−3z)=3(10)+4(6)  ⟹  21x−2y=543(3x - 2y + 4z) + 4(3x + y - 3z) = 3(10) + 4(6) \implies 21x - 2y = 54. Dividing both sides of 21x−2y=5421x - 2y = 54 by 3 directly gives 7x−23y=187x - \frac{2}{3}y = 18.
Estimated Time:2m 0s
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Topics

Systems of Linear Equations

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