Question

1 Solve the equations 5x + 2y = 14, 3x - 4y - 24.
x = 4, y = -4
x = 4, y = -2
x = 4, y = -3
x = 4, y = 3
2 Solve the linear equation 2x+3y=1, 5x+7y=3.
x=2, y=-1
x=4, y=-2
x=2, y=-2
x=5, y=-3
3 Solve the linear equations 2x+4y=10 and 3x+6y=15.
x=5-2a,y=a
x=5-2a,y=4
x=5,y=a
x=-2a,y=a
4 Solve the set of linear equations by the matrix method : a+3b+2c=3 , 2a-b-3c= -8, 5a+2b+c=9. Sove for c
3
1
5
7
5 Solve the linear equation : 2x+3y=3, x-2y=5 and 3x+2y=7.
x=2 and y=-1
x=3 and y=1
x=3 and y=-1
x=1 and y=-1

Answer 1

Answer on Question #55750 – Math – Linear Algebra

1. Solve the equations $5x + 2y = 14$, $3x - 4y = 24$.

$$\begin{array}{l} x = 4, y = -4 \\ x = 4, y = -2 \\ x = 4, y = -3 \\ x = 4, y = 3 \\ \end{array}$$

Solution

Multiply the first equation of the system by 2:

$$\left\{ \begin{array}{l} 5x + 2y = 14 \mid *2 \\ 3x - 4y = 24 \end{array} \right.$$

Add two equations

$$+ \left\{ \begin{array}{l} 10x + 4y = 28 \\ 3x - 4y = 24 \end{array} \right.$$

13x = 52.

Divide both sides of the equation by 13:

$$x = 4.$$

Substitute for $x = 4$ into the first equation of the initial system:

$$5*4 + 2y = 14.$$

Collect similar terms:

$$2y = 14 - 20.$$

Simplify:

$$2y = -6.$$

Divide both sides by 2:

$$y = -3.$$

Answer: $x = 4, y = -3$.

2. Solve the linear equation $2x + 3y = 1$, $5x + 7y = 3$.

$$\begin{array}{l} x = 2, y = -1 \\ x = 4, y = -2 \\ x = 2, y = -2 \\ x = 5, y = -3 \\ \end{array}$$

Solution

Multiply the first equation of the system by 5 and the second equation by (-2)

$$\left\{ \begin{array}{l} 2x + 3y = 1 \mid *5 \\ 5x + 7y = 3. \quad | *(-2) \end{array} \right.$$

Add two equations:

+ $\left\{ \begin{array}{l}10x + 15y = 5\\ -10x - 14y = -6 \end{array} \right.$

y=-1.

Substitute for $y = -1$ into the first equation of the initial system of equations:

$2x + 3^{*}(-1) = 1$

Collect similar terms:

$2x = 4$

Divide both sides by 2:

$x = 2$

Answer: $x = 2, y = -1$

3. Solve the linear equations $2x + 4y = 10$ and $3x + 6y = 15$ .

$x = 5 - 2a,y = a$

$x = 5 - 2a,y = 4$

$x = 5, y = a$

$x = -2a, y = a$

Solution

Divide the first equation by 2 and the second equation by 3:

$\left\{ \begin{array}{l}2x + 4y = 10\ |:2\\ 3x + 6y = 15\ |:3 \end{array} \right.$

$\left\{ \begin{array}{l}x + 2y = 5\\ x + 2y = 5, \end{array} \right.$

We obtain two identical equations, which give the only equation $x + 2y = 5$ , hence $x = 5 - 2y$ .

if $y = a$ then $x = 5 - 2a$

Answer: $x = 5 - 2a, y = a$

4. Solve the set of linear equations by the matrix method:

$a + 3b + 2c = 3$ , $2a - b - 3c = -8$ $5a + 2b + c = 9$ . Solve for c

3157

Solution

We have the matrix equation

AX=B, where

$A = \left( \begin{array}{rrr}1 & 3 & 2\\ 2 & -1 & -3\\ 5 & 2 & 1 \end{array} \right);X = \left( \begin{array}{l}a\\ b\\ c \end{array} \right);B = \left( \begin{array}{l}3\\ -8\\ 9 \end{array} \right)$

Det $A = 1 \times (-1) \times 1 + 3 \times (-3) \times 5 + 2 \times 2 \times 2 - 5 \times (-1) \times 2 - 2 \times (-3) \times 1 - 1 \times 2 \times 3 = -28$

$$\begin{array}{c c c c c c c c} \mathrm {A} 1 1 = (- 1) ^ {1 + 1} & - 1 - 3 & & & & & \\ & 2 & 1 & = 5 & & & \\ \mathrm {A} 2 1 = (- 1) ^ {2 + 1} & \frac {3}{2} & & & & & \\ & 2 & 1 & = 1 & & & \\ \mathrm {A} 3 1 = (- 1) ^ {3 + 1} & \frac {3}{- 1} & & & & & \\ & - 1 & - 3 & = - 7 & & & \end{array} \quad \begin{array}{c c c c c c c c} \mathrm {A} 1 2 = (- 1) ^ {1 + 2} & \frac {2}{5} & - 3 & & & & \\ & 5 & 1 & = - 1 7 & & & \\ \mathrm {A} 2 2 = (- 1) ^ {2 + 2} & \frac {1}{5} & \frac {2}{1} & & & & \\ & 5 & 1 & = - 9 & & & \\ \mathrm {A} 3 2 = (- 1) ^ {3 + 2} & \frac {1}{2} & \frac {2}{- 3} & = 7 & & & \\ & - 3 & & & & & \end{array} \quad \begin{array}{c c c c c c c c} \mathrm {A} 1 3 = (- 1) ^ {1 + 3} & \frac {2}{5} & - 1 & & & & \\ & 5 & 2 & = 1 3 & & & \\ \mathrm {A} 3 3 = (- 1) ^ {3 + 3} & \frac {1}{2} & \frac {3}{- 1} & & & & \\ & 2 & - 7 & & & & \end{array}$$

The inverse of A is

$$A ^ {- 1} = \begin{array}{c c c c} - \frac {1}{2 8} \left( \begin{array}{c c c} 5 & - 1 7 & 9 \\ 1 & - 9 & 1 3 \\ - 7 & 7 & - 7 \end{array} \right) ^ {T} = - \frac {1}{2 8} \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) \end{array}$$

Using the matrix method,

$$X = \begin{array}{c c c c} - \frac {1}{2 8} \left( \begin{array}{c c c} 5 & 1 & - 7 \\ - 1 7 & - 9 & 7 \\ 9 & 1 3 & - 7 \end{array} \right) \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) _ {=} & \left( \begin{array}{c c c} - 5 / 2 8 & - 1 / 2 8 & 1 / 4 \\ 1 7 / 2 8 & 9 / 2 8 & - 1 / 4 \\ - 9 / 2 8 & - 1 3 / 2 8 & 1 / 4 \end{array} \right) \left( \begin{array}{c} 3 \\ - 8 \\ 9 \end{array} \right) _ {=} & \left( \begin{array}{c} - 1 5 / 2 8 + 8 / 2 8 + 9 / 4 \\ 5 1 / 2 8 - 7 2 / 2 8 - 9 / 4 \\ - 2 7 / 2 8 + 1 0 4 / 2 8 + 9 / 4 \end{array} \right) = \left( \begin{array}{c} 2 \\ - 3 \\ 5 \end{array} \right) \end{array}$$

Thus, $a = 2$ ; $b = -3$ ; $c = 5$ .

Answer: $c = 5$

5. Solve the linear equation: $2x + 3y = 3$ , $x - 2y = 5$ and $3x + 2y = 7$ .

$x = 2$ and $y = -1$

$x = 3$ and $y = 1$

$x = 3$ and $y = -1$

$x = 1$ and $y = -1$

Solution

$$\left\{ \begin{array}{c} 2 x + 3 y = 3 \\ x - 2 y = 5 \\ 3 x + 2 y = 7 \end{array} \right.$$

Replace the third equation by the sum of the second and the third equations of the system:

$$\left\{ \begin{array}{l} 2 x + 3 y = 3 \\ x - 2 y = 5 \\ 4 x = 1 2 \end{array} \right.$$

Divide both sides of the third equation by 3:

$x = 3$

Substitute for $x = 3$ into the first and the second equations of the system:

$$\left\{ \begin{array}{l} 6 + 3 y = 3, \\ 3 - 2 y = 5, \\ x = 3, \end{array} \right.$$

Replace the first equation by the sum of the first and the second ones:

$9 + y = 8$ , hence $y = 8 - 9$ , that is, $y = -1$

Notice that $y = -1$ satisfies each equation of the previous system.

Answer: $x = 3$ and $y = -1$ .

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