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{:[3s+2t-3=c],[-7s-5t=4]:} Consider the system of equations, where c is a constant. For which value of c is there exactly one (s,t) solution where s=-1 ? Choose 1 answer: (A) -4.8 (B) -4(4)/(7) (c) 18 (D) None of the above

3s+2t−3=c−7s−5t=4 \begin{array}{r} 3 s+2 t-3=c \\ -7 s-5 t=4 \end{array} \newlineConsider the system of equations, where c c is a constant. For which value of c c is there exactly one (s,t) (s, t) solution where s=−1 s=-1 ?\newlineChoose 11 answer:\newline(A) −4-4.88\newline(B) −447 -4 \frac{4}{7} \newline(C) 1818\newline(D) None of the above

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Q. 3s+2t−3=c−7s−5t=4 \begin{array}{r} 3 s+2 t-3=c \\ -7 s-5 t=4 \end{array} \newlineConsider the system of equations, where c c is a constant. For which value of c c is there exactly one (s,t) (s, t) solution where s=−1 s=-1 ?\newlineChoose 11 answer:\newline(A) −4-4.88\newline(B) −447 -4 \frac{4}{7} \newline(C) 1818\newline(D) None of the above
  1. System of Equations: We have the system of equations:\newline{3s+2t−3=c−7s−5t=4 \begin{cases} 3s + 2t - 3 = c \\ -7s - 5t = 4 \end{cases} \newlineWe need to find the value of c for which there is exactly one solution (s,t) where s = −1-1.
  2. Substituting s = −1-1: First, let's substitute s = −1-1 into the second equation to find the corresponding value of t.\newline−7(−1)−5t=4 -7(-1) - 5t = 4 \newline7−5t=4 7 - 5t = 4
  3. Solving for t: Now, let's solve for t:\newline−5t=4−7 -5t = 4 - 7 \newline−5t=−3 -5t = -3 \newlinet=−3−5 t = \frac{-3}{-5} \newlinet=35 t = \frac{3}{5}
  4. Substituting s = −1-1 and t = 33/55: Next, we substitute s = −1-1 and t = 35\frac{3}{5} into the first equation to find the value of c.\newline3(−1)+2(35)−3=c 3(-1) + 2\left(\frac{3}{5}\right) - 3 = c \newline−3+65−3=c -3 + \frac{6}{5} - 3 = c \newline−155+65−155=c -\frac{15}{5} + \frac{6}{5} - \frac{15}{5} = c \newlinec=−245 c = -\frac{24}{5} \newlinec=−4.8 c = -4.8

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