Solve each system of equations.3a+2b=275a-7b+c=5-2a+10b+5c=-29

Q38. - Algebra 2 | StudyQuestionHub

Solve each system of equations.

\begin{matrix} 3 a + 2 b = 27 \\ 5 a - 7 b + c = 5 \\ - 2 a + 10 b + 5 c = - 29 \end{matrix}

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Answer

The solution of given system of equation is $(7, 3, - 9)$ .

1Step 1 – Description of step.

Multiply $5 a - 7 b + c = 5$ by 5 and subtract $- 2 a + 10 b + 5 c = - 29$ from the obtained equation.

\begin{matrix} 25 a - 35 b + 5 c - (- 2 a + 10 b + 5 c) = 25 - (- 29) \\ 25 a - 35 b + 5 c + 2 a - 10 b - 5 c = 54 \\ 25 a - 35 b + 2 a - 10 b = 54 \\ 27 a - 45 b = 54 \end{matrix}

2Step 2 – Description of step.

Multiply $3 a + 2 b = 27$ by 9 and subtract $27 a - 45 b = 54$ from the obtained equation.

\begin{matrix} 27 a + 18 b - (27 a - 45 b) = 243 - 54 \\ 27 a + 18 b - 27 a + 45 b = 189 \\ \frac{63 b}{63} = \frac{189}{63} \\ b = 3 \end{matrix}

3Step 3 – Description of step.

Substitute 3 for $b$ into $3 a + 2 b = 27$ and simplify to obtain the value of $a$ .

\begin{matrix} 3 a + 2 (3) = 27 \\ 3 a + 6 = 27 \\ \frac{3 a}{3} = \frac{21}{3} \\ a = 7 \end{matrix}

4Step 4 – Description of step.

Substitute 3 for $b$ and 7 for $a$ into $5 a - 7 b + c = 5$ and simplify to obtain the value of $c$ .

\begin{matrix} 5 (7) - 7 (3) + c = 5 \\ 35 - 21 + c = 5 \\ 14 + c = 5 \\ c = - 9 \end{matrix}

Therefore, the solution of given system of equation is $(7, 3, - 9)$ .