Question

1. $-5x - 7y = 19$

$-4x - 5y = 11$

1. $-16x + 14y = -48$

$-8x + 7y = -24$

1. $-5x - 4y = 6$

$8x - 2y = 24$

1. $-4x + 6y = -9$

$-8x + 12y = 5$

Explanation:

Let's solve each system of equations one by one.

Problem 5:

We have the system:

$$\begin{cases} -5x - 7y = 19 \\ -4x - 5y = 11 \end{cases}$$

Step 1: Eliminate one variable. Let's eliminate \(x\). Multiply the first equation by \(4\) and the second equation by \(5\):

First equation multiplied by \(4\): \(4(-5x - 7y)=4\times19\) gives \(-20x - 28y = 76\)
Second equation multiplied by \(5\): \(5(-4x - 5y)=5\times11\) gives \(-20x - 25y = 55\)

Step 2: Subtract the second new equation from the first new equation:

\((-20x - 28y)-(-20x - 25y)=76 - 55\)
Simplify: \(-20x - 28y + 20x + 25y = 21\)
\(-3y = 21\)
Divide both sides by \(-3\): \(y=\frac{21}{-3}=-7\)

Step 3: Substitute \(y = -7\) into one of the original equations, say \(-4x - 5y = 11\):

\(-4x - 5(-7)=11\)
\(-4x + 35 = 11\)
Subtract \(35\) from both sides: \(-4x=11 - 35=-24\)
Divide by \(-4\): \(x=\frac{-24}{-4}=6\)

So the solution for problem 5 is \(x = 6,y=-7\)

Problem 6:

We have the system:

$$\begin{cases} -16x + 14y = -48 \\ -8x + 7y = -24 \end{cases}$$

Step 1: Notice that the first equation is \(2\) times the second equation. Multiply the second equation by \(2\): \(2(-8x + 7y)=2\times(-24)\) gives \(-16x + 14y=-48\), which is the same as the first equation.

This means the two equations are dependent (they represent the same line), so there are infinitely many solutions. The solution can be expressed in terms of \(x\) or \(y\). Let's solve for \(y\) from the second equation:
\(-8x + 7y=-24\)
\(7y=8x - 24\)
\(y=\frac{8x - 24}{7}\) where \(x\) is any real number.

Problem 7:

We have the system:

$$\begin{cases} -5x - 4y = 6 \\ 8x - 2y = 24 \end{cases}$$

Step 1: Eliminate \(y\). Multiply the second equation by \(2\): \(2(8x - 2y)=2\times24\) gives \(16x - 4y = 48\)

Step 2: Subtract the first equation from this new equation:

\((16x - 4y)-(-5x - 4y)=48 - 6\)
Simplify: \(16x - 4y + 5x + 4y = 42\)
\(21x = 42\)
Divide by \(21\): \(x = \frac{42}{21}=2\)

Step 3: Substitute \(x = 2\) into the second original equation \(8x - 2y = 24\):

\(8(2)-2y=24\)
\(16 - 2y = 24\)
Subtract \(16\) from both sides: \(-2y=24 - 16 = 8\)
Divide by \(-2\): \(y=\frac{8}{-2}=-4\)

So the solution for problem 7 is \(x = 2,y=-4\)

Problem 8:

We have the system:

$$\begin{cases} -4x + 6y = -9 \\ -8x + 12y = 5 \end{cases}$$

Step 1: Multiply the first equation by \(2\): \(2(-4x + 6y)=2\times(-9)\) gives \(-8x + 12y=-18\)

Step 2: Now we have two equations:

\(-8x + 12y=-18\) (from first equation multiplied by \(2\))
\(-8x + 12y = 5\) (second equation)

Subtract the first new equation from the second equation:
\((-8x + 12y)-(-8x + 12y)=5-(-18)\)
Simplify: \(0=23\), which is a contradiction.

This means the system has no solution (the two lines are parallel and never intersect).

Final Answers:

1. \(x = 6,y=-7\)
2. Infinitely many solutions (\(y=\frac{8x - 24}{7},x\in\mathbb{R}\))
3. \(x = 2,y=-4\)
4. No solution

Answer:

Let's solve each system of equations one by one.

Problem 5:

We have the system:

$$\begin{cases} -5x - 7y = 19 \\ -4x - 5y = 11 \end{cases}$$

Step 1: Eliminate one variable. Let's eliminate \(x\). Multiply the first equation by \(4\) and the second equation by \(5\):

First equation multiplied by \(4\): \(4(-5x - 7y)=4\times19\) gives \(-20x - 28y = 76\)
Second equation multiplied by \(5\): \(5(-4x - 5y)=5\times11\) gives \(-20x - 25y = 55\)

Step 2: Subtract the second new equation from the first new equation:

\((-20x - 28y)-(-20x - 25y)=76 - 55\)
Simplify: \(-20x - 28y + 20x + 25y = 21\)
\(-3y = 21\)
Divide both sides by \(-3\): \(y=\frac{21}{-3}=-7\)

Step 3: Substitute \(y = -7\) into one of the original equations, say \(-4x - 5y = 11\):

\(-4x - 5(-7)=11\)
\(-4x + 35 = 11\)
Subtract \(35\) from both sides: \(-4x=11 - 35=-24\)
Divide by \(-4\): \(x=\frac{-24}{-4}=6\)

So the solution for problem 5 is \(x = 6,y=-7\)

Problem 6:

We have the system:

$$\begin{cases} -16x + 14y = -48 \\ -8x + 7y = -24 \end{cases}$$

Step 1: Notice that the first equation is \(2\) times the second equation. Multiply the second equation by \(2\): \(2(-8x + 7y)=2\times(-24)\) gives \(-16x + 14y=-48\), which is the same as the first equation.

This means the two equations are dependent (they represent the same line), so there are infinitely many solutions. The solution can be expressed in terms of \(x\) or \(y\). Let's solve for \(y\) from the second equation:
\(-8x + 7y=-24\)
\(7y=8x - 24\)
\(y=\frac{8x - 24}{7}\) where \(x\) is any real number.

Problem 7:

We have the system:

$$\begin{cases} -5x - 4y = 6 \\ 8x - 2y = 24 \end{cases}$$

Step 1: Eliminate \(y\). Multiply the second equation by \(2\): \(2(8x - 2y)=2\times24\) gives \(16x - 4y = 48\)

Step 2: Subtract the first equation from this new equation:

\((16x - 4y)-(-5x - 4y)=48 - 6\)
Simplify: \(16x - 4y + 5x + 4y = 42\)
\(21x = 42\)
Divide by \(21\): \(x = \frac{42}{21}=2\)

Step 3: Substitute \(x = 2\) into the second original equation \(8x - 2y = 24\):

\(8(2)-2y=24\)
\(16 - 2y = 24\)
Subtract \(16\) from both sides: \(-2y=24 - 16 = 8\)
Divide by \(-2\): \(y=\frac{8}{-2}=-4\)

So the solution for problem 7 is \(x = 2,y=-4\)

Problem 8:

We have the system:

$$\begin{cases} -4x + 6y = -9 \\ -8x + 12y = 5 \end{cases}$$

Step 1: Multiply the first equation by \(2\): \(2(-4x + 6y)=2\times(-9)\) gives \(-8x + 12y=-18\)

Step 2: Now we have two equations:

\(-8x + 12y=-18\) (from first equation multiplied by \(2\))
\(-8x + 12y = 5\) (second equation)

Subtract the first new equation from the second equation:
\((-8x + 12y)-(-8x + 12y)=5-(-18)\)
Simplify: \(0=23\), which is a contradiction.

This means the system has no solution (the two lines are parallel and never intersect).

Final Answers:

1. \(x = 6,y=-7\)
2. Infinitely many solutions (\(y=\frac{8x - 24}{7},x\in\mathbb{R}\))
3. \(x = 2,y=-4\)
4. No solution