Question

Solve the following system of equations:

x − 2y = 14
x + 3y = 9

(1, 12)

(−1, −12)

(12, −1)

(12, 1)

Answer 1

12, -1

Answer 2

12, −1)

Answer 3

Could any of you too by chance show me the work because I'm still not getting it. thanks

Answer 4

Row reduce this matrix\[ \left[ \begin{array}{ccc}
1 &-2 & 14 \\
1 & 3 & 9 \\ \end{array} \right]\] to get \[ \left[ \begin{array}{ccc}
1 &0 & 12 \\
0 & 1 & -1 \\ \end{array} \right]\]
Thus x=12 and y=-1

or
\[x=\frac{\left|\begin{array}{ccc}
14 & -2 \\
9 & 3 \\ \end{array} \right|}{\left|\begin{array}{ccc}
1& -2 \\
1 & 3 \\ \end{array} \right|}=12\]

\[y=\frac{\left|\begin{array}{ccc}
1 & 14 \\
1 & 9\\ \end{array} \right|}{\left|\begin{array}{ccc}
1& -2 \\
1 & 3 \\ \end{array} \right|}=-1\]

Answer 5

@cyter:I believe that is also known as cramer's rule ( http://en.wikipedia.org/wiki/Cramer%27s_rule)

Answer 6

x - 2y = 14................(a)
x + 3y = 9.................(b)
from (a) x = 14 + 2y...... substitute in b
14 + 2y + 3y =9
5y = -5
y = -1 then substitute in a
x - 2(-1) = 14
x = 14 - 2
x = 12
ans: (12, -1)

Answer 7

Lets use the matrix method:

\[ Ax=B \Rightarrow x=A^{1}B \]

Here,

\[ A = \left[ \begin{array}{ccc}
1 &-2 \\
1 & 3 \\ \end{array} \right] \]

\[ B = \left[ \begin{array}{ccc}
14 \\
9 \\ \end{array} \right] \]

\[ x = \left[ \begin{array}{ccc}
x_1 \\
x_2 \\ \end{array} \right] \]

Again,\[ A^{-1} = \frac15 \left[ \begin{array}{ccccc}
3 &2 \\
-1 & -1\\ \end{array} \right] \]

Giving \[ \left[ \begin{array}{ccc}
x_1 \\
x_2 \\ \end{array} \right] =\left[ \begin{array}{ccc}
12 \\
-1 \\ \end{array} \right] \]

and we are done here :)

Answer 8

or graph \[y = \frac{1}{2}x-7\]
and \[y = -\frac{1}{3}x+3\]
and find the intersection