Q23. Solve each system of equations聽... [FREE SOLUTION]

Chapter 3: Q23. (page 143)

Solve each system of equations
$\begin{matrix} 2 a 鈭� b + 3 c = 鈭� 7 \\ 4 a + 5 b + c = 29 \\ a 鈭� \frac{2}{3} b + \frac{1}{4} c = 鈭� 10 \end{matrix}$

Short Answer

Expert verified

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

Step by step solution

– 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 $4 a + 5 b + c = 29$ .

$\begin{array}{l} \underline{\begin{array}{l} 2 a 鈭� 鈥夆赌夆赌� b + 3 c = 鈭� 鈥夆赌� 7 \\ 4 a + 5 b + 鈥夆赌夆赌� c = 鈥夆赌夆赌� 29 \end{array}} 鈥夆赌夆赌夆€� \begin{matrix} multiplyby5 鈫� \end{matrix} 鈥夆赌� \underline{\begin{array}{l} 10 a 鈭� 5 b + 15 c = 鈭� 35 \\ 鈥夆赌� 4 a + 5 b + 鈥夆赌夆赌夆€� c = 鈥夆赌夆赌� 29 \end{array}} \\ 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌� 14 a + 鈥夆赌� 0 + 16 c = 鈭� 鈥夆赌� 6 \end{array}$

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 $a - \frac{2}{3} b + \frac{1}{4} c = - 10$

$\begin{array}{l} \underline{\begin{array}{l} 2 a 鈭� 鈥夆赌夆赌夆€� b + 3 c = 鈭� 鈥夆赌� 7 \\ 鈥夆赌� a 鈭� \frac{2}{3} b + \frac{1}{4} c = 鈭� 10 \end{array}} 鈥夆赌夆赌夆€� \begin{matrix} multiplyby \frac{2}{3} 鈫� \end{matrix} 鈥夆赌� \underline{\begin{array}{l} 鈥夆赌夆赌夆€� \frac{4}{3} a 鈭� \frac{2}{3} b + 2 c = 鈭� \frac{14}{3} \\ (鈭�) 鈥夆赌夆赌� a 鈭� \frac{2}{3} b + \frac{1}{4} c = 鈭� 10 \end{array}} \\ 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌� \frac{1}{3} a + 鈥夆赌夆赌� 0 + \frac{7}{4} c = 鈥夆赌� \frac{16}{3} \end{array}$

.So, the resultant equation is

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

– 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 $14 a + 16 c = - 6$ .

$\begin{array}{l} \underline{\begin{array}{l} \frac{1}{3} a + \frac{7}{4} c = \frac{16}{3} \\ 14 a + 16 c = 鈭� 6 \end{array}} 鈥夆赌夆赌夆€� \begin{matrix} multiplyby 42 \end{matrix} 鈥夆赌� \underline{\begin{array}{l} 鈥夆赌夆赌夆€夆赌� 14 a + \frac{147}{2} c = 鈥夆赌� 224 \\ (鈭�) 14 a + 鈥夆赌夆赌� 16 c = 鈭� 鈥夆赌夆赌夆€� 6 \end{array}} \\ 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌� 0 + \frac{115}{2} c = 鈥夆赌夆赌� 230 \end{array}$

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

$\begin{matrix} \frac{115}{2} c = 230 \\ \frac{2}{115} 鈰� \frac{115 c}{2} = \frac{2}{115} 鈰� 230 鈥夆赌夆赌夆€夆赌塎ultiplybothsidesby \frac{2}{115} \\ c = 4 \end{matrix}$

– Find the values of a and .b

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

$\begin{matrix} 14 a + 16 c = 鈭� 6 \\ 14 a + 16 (4) = 鈭� 6 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€塖ubstitute 4 for c \\ 14 a + 64 = 鈭� 6 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€塖implify \\ 14 a = 鈭� 70 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塖ubtract 64 frombothsides \\ a = 鈭� 5 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€塂ividebothsidesby14 \end{matrix}$

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

$\begin{matrix} 2 a 鈭� b + 3 c = 鈭� 7 \\ 2 (鈭� 5) 鈭� b + 3 (4) = 鈭� 7 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塻ubstitute 鈭� 5 for a, 4 for c \\ 鈭� 10 鈭� b + 12 = 鈭� 7 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塻implify \\ 2 鈭� b = 鈭� 7 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌塻implify \\ 鈭� b = 鈭� 9 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€塖ubtract 2 鈥塮谤辞尘产辞迟丑蝉颈诲别蝉 \\ b = 9 鈥夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌夆€夆赌夆赌塪ividebothsidesby 鈭� 1 \end{matrix}$

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

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Solve each system of equations
$\begin{matrix} 2 a 鈭� b + 3 c = 鈭� 7 \\ 4 a + 5 b + c = 29 \\ a 鈭� \frac{2}{3} b + \frac{1}{4} c = 鈭� 10 \end{matrix}$

Short Answer

Step by step solution

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

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

– Find the values of a and .b

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91影视

Solve each system of equations2a鈭�b+3c=鈭�74a+5b+c=29a鈭�23b+14c=鈭�10

Short Answer

Step by step solution

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

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

– Find the values of a and .b

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Discrete Mathematics

Calculus

Pure Maths

Probability and Statistics

Applied Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Solve each system of equations
$\begin{matrix} 2 a 鈭� b + 3 c = 鈭� 7 \\ 4 a + 5 b + c = 29 \\ a 鈭� \frac{2}{3} b + \frac{1}{4} c = 鈭� 10 \end{matrix}$