Question 460 of 472MediumMCQ
If the system of linear equations :
\[x+y+2z=62x+3y+az=a+1−x−3y+bz=2 b\]
where a,b∈R, has infinitely many solutions, then 7a+3b is equal to :
Solutions · 1
Official solution
We begin with the system:
\[ x+y+2z2x+3y+az−x−3y+bz=6,=a+1,=2b. \]
Solve the first equation for x: 1. Solve the first equation for x:
x=6−y−2z.
Substitute x=6−y−2z into the second equation: 2. Substitute x=6−y−2z into the second equation:
2(6−y−2z)+3y+az=a+1.
Expanding and simplifying:
\[ 12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11. \]
Call this Equation (I).
Substitute x=6−y−2z into the third equation: 3. Substitute x=6−y−2z into the third equation:
−(6−y−2z)−3y+bz=2b.
Expanding and simplifying:
\[ -6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6. \]
Call this Equation (II).
For the system to have infinitely many solutions, the two equations in y and z must be dependent---that is, one must be a constant multiple of the other. Assume there exists a constant k such tha: 4. For the system to have infinitely many solutions, the two equations in y and z must be dependent---that is, one must be a constant multiple of the other. Assume there exists a constant k such that
\[ -2 = k \cdot 1 \quad \Longrightarrow \quad k = -2. \]
Apply this to the coefficient of z and the constant term.
For the z-coefficient in Equations (I) and (II):
b+2=k(a−4)=−2(a−4)=−2a+8.
Thus,
b=−2a+6.
For the constant term:
2b+6=k(a−11)=−2(a−11)=−2a+22.
Substitute b=−2a+6 into this equation:
\[ 2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22. \]
Simplify:
−4a+18=−2a+22.
Solve for a:
\[ -4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22, \]
\[ 2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2. \]
Substitute a=−2 into b=−2a+6:
b=−2(−2)+6=4+6=10.
We now comput: 5. We now compute
\[ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. \]
Thus, the value of 7a+3b is
\[ \boxed{16}. \]
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