Question

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

Answer 1

another similiar question

Answer 2

can you explain simply to me

Answer 3

give me a similiar problem with the answer

Answer 4

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 5

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 6

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 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!