i need a similiar problem to solve a linear system algebraically similiar to y=2x+4 3x+y=9

Answer 1

To create a similar problem involving a linear system that can be solved algebraically, we'll maintain similar structures and relationships.

Here's a new system of equations:

1. \( y = 3x + 1 \)
2. \( 4x + y = 13 \)

To solve this system algebraically, you can follow these steps:

1. Substitute \(y\) from the first equation into the second equation.
2. Solve for \(x\).
3. Substitute the value of \(x\) back into the first equation to find \(y\).

Let's break it down:

Step 1: Substitute \(y\)

From the first equation, we have:

\[ y = 3x + 1 \]

Substituting \(y\) in the second equation:

\[ 4x + (3x + 1) = 13 \]

Step 2: Solve for \(x\)

Combine like terms:

\[ 4x + 3x + 1 = 13 \] \[ 7x + 1 = 13 \]

Subtract \(1\) from both sides:

\[ 7x = 12 \]

Now, divide by \(7\):

\[ x = \frac{12}{7} \]

Step 3: Solve for \(y\)

Now, substitute \(x\) back into the first equation:

\[ y = 3\left(\frac{12}{7}\right) + 1 \] \[ y = \frac{36}{7} + 1 = \frac{36}{7} + \frac{7}{7} = \frac{43}{7} \]

Solution

The solution to the system of equations is:

\[ x = \frac{12}{7}, \quad y = \frac{43}{7} \]

This new system mirrors the structure of your original problem, allowing for similar methods of algebraic solution.

Answer 2

another similiar question

Answer 3

Sure! Here's another similar problem involving a linear system:

1. \( y = -\frac{1}{2}x + 5 \)
2. \( 2x - 3y = -6 \)

Steps to Solve:

Just like before, we can solve this system algebraically.

Step 1: Substitute \(y\)

From the first equation, we have:

\[ y = -\frac{1}{2}x + 5 \]

Substituting \(y\) into the second equation:

\[ 2x - 3\left(-\frac{1}{2}x + 5\right) = -6 \]

Step 2: Solve for \(x\)

Distributing the \(-3\):

\[ 2x + \frac{3}{2}x - 15 = -6 \]

Combine like terms:

\[ 2x + \frac{3}{2}x = \frac{4}{2}x + \frac{3}{2}x = \frac{7}{2}x \]

So, we have:

\[ \frac{7}{2}x - 15 = -6 \]

Adding 15 to both sides:

\[ \frac{7}{2}x = 9 \]

Now, multiply both sides by \(\frac{2}{7}\):

\[ x = \frac{18}{7} \]

Step 3: Solve for \(y\)

Now substitute \(x\) back into the first equation:

\[ y = -\frac{1}{2}\left(\frac{18}{7}\right) + 5 \]

Calculating,

\[ y = -\frac{9}{7} + 5 = -\frac{9}{7} + \frac{35}{7} = \frac{26}{7} \]

Solution

The solution to this system of equations is:

\[ x = \frac{18}{7}, \quad y = \frac{26}{7} \]

Feel free to solve this system, and let me know if you need more problems or further assistance!

Answer 4

can you explain simply to me

Answer 5

Of course! Let's break it down step-by-step in a simpler way using the new system of equations.

The System of Equations

1. \( y = -\frac{1}{2}x + 5 \) (This is the first equation.)
2. \( 2x - 3y = -6 \) (This is the second equation.)

We want to find the values of \(x\) and \(y\) that make both equations true.

Step 1: Substitute \(y\)

Since we already have \(y\) expressed in terms of \(x\) from the first equation, we can use that to find \(x\).

• We take the first equation and plug it into the second equation.

\[ y = -\frac{1}{2}x + 5 \]

Now substitute this \(y\) value into the second equation:

\[ 2x - 3\left(-\frac{1}{2}x + 5\right) = -6 \]

Step 2: Simplify the Second Equation

Now, let's simplify the left-hand side:

1. Distribute \(-3\) in \( -3\left(-\frac{1}{2}x + 5\right)\):

   \[ -3 \times -\frac{1}{2}x = \frac{3}{2}x \]

   \[ -3 \times 5 = -15 \]

So, we can rewrite the equation as:

\[ 2x + \frac{3}{2}x - 15 = -6 \]

Step 3: Combine Like Terms

Next, combine the \(x\) terms:

\[ 2x + \frac{3}{2}x = \frac{4}{2}x + \frac{3}{2}x = \frac{7}{2}x \]

So now we have:

\[ \frac{7}{2}x - 15 = -6 \]

Step 4: Solve for \(x\)

Now, let's isolate \(x\):

1. Add 15 to both sides:

   \[ \frac{7}{2}x = -6 + 15 \]

   \[ \frac{7}{2}x = 9 \]

2. Now, multiply both sides by \(\frac{2}{7}\) (which is the same as dividing by \(\frac{7}{2}\)):

   \[ x = 9 \cdot \frac{2}{7} = \frac{18}{7} \]

Step 5: Find \(y\)

Now that we have \(x\), we can find \(y\). We use the first equation again:

\[ y = -\frac{1}{2}x + 5 \]

Substituting \(x = \frac{18}{7}\):

\[ y = -\frac{1}{2}\left(\frac{18}{7}\right) + 5 \]

Calculating this:

\[ y = -\frac{9}{7} + 5 \]

Convert 5 into a fraction:

\[ 5 = \frac{35}{7} \]

So:

\[ y = -\frac{9}{7} + \frac{35}{7} = \frac{26}{7} \]

Final Solution

The final solution to the system is:

\[ x = \frac{18}{7} \quad \text{and} \quad y = \frac{26}{7} \]

And that’s it! We have found the values for \(x\) and \(y\) that satisfy both equations. If you have any more questions, feel free to ask!

Answer 6

give me a similiar problem with the answer

Answer 7

Sure! Here’s another similar problem with its solution provided:

New System of Equations

1. \( y = 4x - 3 \)
2. \( 5x + 2y = 14 \)

Solution Steps:

Step 1: Substitute \(y\)

We start with the first equation:

\[ y = 4x - 3 \]

Now we substitute this expression for \(y\) into the second equation:

\[ 5x + 2(4x - 3) = 14 \]

Step 2: Simplify the Second Equation

Distributing \(2\) gives:

\[ 5x + 8x - 6 = 14 \]

Combine like terms:

\[ 13x - 6 = 14 \]

Step 3: Solve for \(x\)

Add \(6\) to both sides:

\[ 13x = 20 \]

Now divide by \(13\):

\[ x = \frac{20}{13} \]

Step 4: Find \(y\)

Now substitute \(x = \frac{20}{13}\) back into the first equation to find \(y\):

\[ y = 4\left(\frac{20}{13}\right) - 3 \]

Calculating this:

\[ y = \frac{80}{13} - 3 \]

Convert \(3\) into a fraction:

\[ 3 = \frac{39}{13} \]

So:

\[ y = \frac{80}{13} - \frac{39}{13} = \frac{41}{13} \]

Final Solution

The solution to the system of equations is:

\[ x = \frac{20}{13}, \quad y = \frac{41}{13} \]

If you have any further questions or want additional problems, just let me know!