Question 50 of 472MediumMCQ
If the system of linear equations :
x+y+2z=62x+3y+az=a+1−x−3y+bz=2b
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+2z=6,2x+3y+az=a+1,−x−3y+bz=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⟹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⟹−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⋅1⟹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⟹−4a+12+6=−2a+22.
Simplify:
−4a+18=−2a+22.
Solve for a:
−4a+18+4a=−2a+22+4a⟹18=2a+22,
2a=18−22=−4⟹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
16.
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