Solve each system of equations 2a−b+3c=−74a+5b+c=29a−23b+14c=−10

Q23. - Algebra 2 | StudyQuestionHub

Q23.

Question

Solve each system of equations

$\begin{matrix} 2 a - b + 3 c = - 7 \\ 4 a + 5 b + c = 29 \\ a - \frac{2}{3} b + \frac{1}{4} c = - 10 \end{matrix}$

Step-by-Step Solution

Verified

Answer

The solution set of the given system of equations is . $(- 5, 9, 4)$

1Step 1 – Use the elimination method to get the system of equations in two variables.

Multiply the equation $2 a - b + 3 c = - 7$ by $5$ and add the new resultant equation to the equation $4 a + 5 b + c = 29$ .

$\begin{array}{l} \underline{\begin{array}{l} 2 a - b + 3 c = - 7 \\ 4 a + 5 b + c = 29 \end{array}} \begin{matrix} multiply by 5 \to \end{matrix} \underline{\begin{array}{l} 10 a - 5 b + 15 c = - 35 \\ 4 a + 5 b + c = 29 \end{array}} \\ 14 a + 0 + 16 c = - 6 \end{array}$

So, the resultant equation is $14 a + 16 c = - 6$ .

Multiply the equation $2 a - b + 3 c = - 7$ by $\frac{2}{3}$ and subtract the new resultant equation from the equation $a - \frac{2}{3} b + \frac{1}{4} c = - 10$

$\begin{array}{l} \underline{\begin{array}{l} 2 a - b + 3 c = - 7 \\ a - \frac{2}{3} b + \frac{1}{4} c = - 10 \end{array}} \begin{matrix} multiply by \frac{2}{3} \to \end{matrix} \underline{\begin{array}{l} \frac{4}{3} a - \frac{2}{3} b + 2 c = - \frac{14}{3} \\ (-) a - \frac{2}{3} b + \frac{1}{4} c = - 10 \end{array}} \\ \frac{1}{3} a + 0 + \frac{7}{4} c = \frac{16}{3} \end{array}$

.So, the resultant equation is

\frac{1}{3} a + \frac{7}{4} c = \frac{16}{3}

2Step 2 – Use the elimination method to solve the system of two equations.

Multiply $\frac{1}{3} a + \frac{7}{4} c = \frac{16}{3}$ by 42 and subtract the new resultant equation to $14 a + 16 c = - 6$ .

$\begin{array}{l} \underline{\begin{array}{l} \frac{1}{3} a + \frac{7}{4} c = \frac{16}{3} \\ 14 a + 16 c = - 6 \end{array}} \begin{matrix} multiply by 42 \end{matrix} \underline{\begin{array}{l} 14 a + \frac{147}{2} c = 224 \\ (-) 14 a + 16 c = - 6 \end{array}} \\ 0 + \frac{115}{2} c = 230 \end{array}$

Solve $\frac{115}{2} c = 230$ for c :

$\begin{matrix} \frac{115}{2} c = 230 \\ \frac{2}{115} \cdot \frac{115 c}{2} = \frac{2}{115} \cdot 230 Multiply both sides by \frac{2}{115} \\ c = 4 \end{matrix}$

3Step 3 – Find the values of a and .b

Substitute $c = 4 i n 14 a + 16 c = - 6$ and find the value of $a$

$\begin{matrix} 14 a + 16 c = - 6 \\ 14 a + 16 (4) = - 6 Substitute 4 for c \\ 14 a + 64 = - 6 Simplify \\ 14 a = - 70 Subtract 64 from both sides \\ a = - 5 Divide both sides by 14 \end{matrix}$

Substitute $a = - 5, c = 4 i n 2 a - b + 3 c = - 7$ and find the value of $b$

$\begin{matrix} 2 a - b + 3 c = - 7 \\ 2 (- 5) - b + 3 (4) = - 7 substitute - 5 for a, 4 for c \\ - 10 - b + 12 = - 7 simplify \\ 2 - b = - 7 simplify \\ - b = - 9 Subtract 2 from both sides \\ b = 9 divide both sides by - 1 \end{matrix}$

Hence, the solution of the given system of equations is $(a, b, c) = (- 5, 9, 4)$ .

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