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Consider the system of equations. Which of the following statements is true?\newlineChoose 11 answer:\newline(A) There is only one solution (e,f)(e,f) and e⋅fe\cdot f is positive.\newline(B) There is only one solution (e,f)(e,f) and e⋅fe\cdot f is negative.\newline(C) There are infinitely many solutions.\newline(D) There are no solutions.

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Q. Consider the system of equations. Which of the following statements is true?\newlineChoose 11 answer:\newline(A) There is only one solution (e,f)(e,f) and e⋅fe\cdot f is positive.\newline(B) There is only one solution (e,f)(e,f) and e⋅fe\cdot f is negative.\newline(C) There are infinitely many solutions.\newline(D) There are no solutions.
  1. Simplify first equation: First, let's simplify the first equation by distributing the 55 on the right side.11e−9f+1=5f−15e11e - 9f + 1 = 5f - 15eNow, let's combine like terms.11e+15e=5f+9f+111e + 15e = 5f + 9f + 126e=14f+126e = 14f + 1
  2. Combine like terms: Next, we can express ff in terms of ee using the second equation.2f+4e=32f + 4e = 32f=3−4e2f = 3 - 4ef=3−4e2f = \frac{3 - 4e}{2}
  3. Express ff in terms of ee: Now, let's substitute the expression for ff from the second equation into the first equation.26e=14(3−4e2)+126e = 14\left(\frac{3 - 4e}{2}\right) + 126e=7(3−4e)+126e = 7(3 - 4e) + 126e=21−28e+126e = 21 - 28e + 126e+28e=2226e + 28e = 2254e=2254e = 22e=2254e = \frac{22}{54}e=1127e = \frac{11}{27}
  4. Substitute ff into first equation: With the value of ee found, we can now solve for ff.
    f=3−4(1127)2f = \frac{3 - 4(\frac{11}{27})}{2}
    f=3−44272f = \frac{3 - \frac{44}{27}}{2}
    f=81−44542f = \frac{\frac{81 - 44}{54}}{2}
    f=3754/2f = \frac{37}{54} / 2
    f=37108f = \frac{37}{108}
    f=37108f = \frac{37}{108}
  5. Solve for ee: Now that we have the values of ee and ff, we can check if the product e⋅fe\cdot f is positive or negative.\newlinee⋅f=(1127)⋅(37108)e\cdot f = \left(\frac{11}{27}\right) \cdot \left(\frac{37}{108}\right)\newlinee⋅f=4072916e\cdot f = \frac{407}{2916}\newlineSince both ee and ff are positive, their product is also positive.
  6. Solve for ff: Finally, we need to verify that this solution satisfies both original equations.\newlineSubstitute e=1127e = \frac{11}{27} and f=37108f = \frac{37}{108} into the second equation:\newline2(37108)+4(1127)=32\left(\frac{37}{108}\right) + 4\left(\frac{11}{27}\right) = 3\newline74108+4427=3\frac{74}{108} + \frac{44}{27} = 3\newline(74+2×44)/108=3\left(74 + 2\times44\right) / 108 = 3\newline(74+88)/108=3\left(74 + 88\right) / 108 = 3\newline162108=3\frac{162}{108} = 3\newline3=33 = 3\newlineThis confirms that the solution satisfies the second equation.
  7. Check sign of e⋅fe\cdot f: Now, let's check the first equation with the found values of ee and ff.
    11(1127)−9(37108)+1=5(37108−3(1127))11(\frac{11}{27}) - 9(\frac{37}{108}) + 1 = 5(\frac{37}{108} - 3(\frac{11}{27}))
    12127−333108+1=185108−15(1127)\frac{121}{27} - \frac{333}{108} + 1 = \frac{185}{108} - 15(\frac{11}{27})
    (121⋅4−333+108⋅4)/108=(185−15⋅44)/108(121\cdot 4 - 333 + 108\cdot 4) / 108 = (185 - 15\cdot 44) / 108
    (484−333+432)/108=(185−660)/108(484 - 333 + 432) / 108 = (185 - 660) / 108
    583108=−475108\frac{583}{108} = -\frac{475}{108}
    583108=−475108\frac{583}{108} = -\frac{475}{108}
    This confirms that the solution satisfies the first equation.

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