Answer :
Certainly! Let's solve the given system of equations step by step.
We have the following system of equations:
1. [tex]\( 3xy - y^2 = 8 \)[/tex]
2. [tex]\( x - 2y = 1 \)[/tex]
### Step 1: Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] using the second equation
Starting with the second equation:
[tex]\[ x - 2y = 1 \][/tex]
We can solve for [tex]\( x \)[/tex] as follows:
[tex]\[ x = 1 + 2y \][/tex]
### Step 2: Substitute [tex]\( x \)[/tex] in the first equation
Now we substitute [tex]\( x = 1 + 2y \)[/tex] into the first equation:
[tex]\[ 3(1 + 2y)y - y^2 = 8 \][/tex]
Distribute [tex]\( y \)[/tex] through the parentheses:
[tex]\[ 3y + 6y^2 - y^2 = 8 \][/tex]
Combine like terms:
[tex]\[ 3y + 5y^2 = 8 \][/tex]
### Step 3: Form a quadratic equation
We have the quadratic equation:
[tex]\[ 5y^2 + 3y - 8 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To find the values of [tex]\( y \)[/tex], we can use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 5 \cdot (-8)}}{2 \cdot 5} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{9 + 160}}{10} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{169}}{10} \][/tex]
[tex]\[ y = \frac{-3 \pm 13}{10} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-3 + 13}{10} = \frac{10}{10} = 1 \][/tex]
[tex]\[ y = \frac{-3 - 13}{10} = \frac{-16}{10} = -\frac{8}{5} \][/tex]
### Step 5: Find corresponding [tex]\( x \)[/tex] values
Now, substitute each value of [tex]\( y \)[/tex] back into the expression [tex]\( x = 1 + 2y \)[/tex] to find the corresponding [tex]\( x \)[/tex] values:
For [tex]\( y = 1 \)[/tex]:
[tex]\[ x = 1 + 2 \cdot 1 = 1 + 2 = 3 \][/tex]
For [tex]\( y = -\frac{8}{5} \)[/tex]:
[tex]\[ x = 1 + 2 \cdot \left(-\frac{8}{5}\right) = 1 - \frac{16}{5} = \frac{5}{5} - \frac{16}{5} = -\frac{11}{5} \][/tex]
### Step 6: Write the solution set
The solutions to the system of equations are:
[tex]\[ (x, y) = \left(3, 1\right) \][/tex]
[tex]\[ (x, y) = \left(-\frac{11}{5}, -\frac{8}{5}\right) \][/tex]
Thus, the solution set is:
[tex]\[ \boxed{\left\{\left(3, 1\right), \left(-\frac{11}{5}, -\frac{8}{5}\right)\right\}} \][/tex]
We have the following system of equations:
1. [tex]\( 3xy - y^2 = 8 \)[/tex]
2. [tex]\( x - 2y = 1 \)[/tex]
### Step 1: Express [tex]\( x \)[/tex] in terms of [tex]\( y \)[/tex] using the second equation
Starting with the second equation:
[tex]\[ x - 2y = 1 \][/tex]
We can solve for [tex]\( x \)[/tex] as follows:
[tex]\[ x = 1 + 2y \][/tex]
### Step 2: Substitute [tex]\( x \)[/tex] in the first equation
Now we substitute [tex]\( x = 1 + 2y \)[/tex] into the first equation:
[tex]\[ 3(1 + 2y)y - y^2 = 8 \][/tex]
Distribute [tex]\( y \)[/tex] through the parentheses:
[tex]\[ 3y + 6y^2 - y^2 = 8 \][/tex]
Combine like terms:
[tex]\[ 3y + 5y^2 = 8 \][/tex]
### Step 3: Form a quadratic equation
We have the quadratic equation:
[tex]\[ 5y^2 + 3y - 8 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To find the values of [tex]\( y \)[/tex], we can use the quadratic formula [tex]\( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 5 \)[/tex], [tex]\( b = 3 \)[/tex], and [tex]\( c = -8 \)[/tex]:
[tex]\[ y = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 5 \cdot (-8)}}{2 \cdot 5} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{9 + 160}}{10} \][/tex]
[tex]\[ y = \frac{-3 \pm \sqrt{169}}{10} \][/tex]
[tex]\[ y = \frac{-3 \pm 13}{10} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{-3 + 13}{10} = \frac{10}{10} = 1 \][/tex]
[tex]\[ y = \frac{-3 - 13}{10} = \frac{-16}{10} = -\frac{8}{5} \][/tex]
### Step 5: Find corresponding [tex]\( x \)[/tex] values
Now, substitute each value of [tex]\( y \)[/tex] back into the expression [tex]\( x = 1 + 2y \)[/tex] to find the corresponding [tex]\( x \)[/tex] values:
For [tex]\( y = 1 \)[/tex]:
[tex]\[ x = 1 + 2 \cdot 1 = 1 + 2 = 3 \][/tex]
For [tex]\( y = -\frac{8}{5} \)[/tex]:
[tex]\[ x = 1 + 2 \cdot \left(-\frac{8}{5}\right) = 1 - \frac{16}{5} = \frac{5}{5} - \frac{16}{5} = -\frac{11}{5} \][/tex]
### Step 6: Write the solution set
The solutions to the system of equations are:
[tex]\[ (x, y) = \left(3, 1\right) \][/tex]
[tex]\[ (x, y) = \left(-\frac{11}{5}, -\frac{8}{5}\right) \][/tex]
Thus, the solution set is:
[tex]\[ \boxed{\left\{\left(3, 1\right), \left(-\frac{11}{5}, -\frac{8}{5}\right)\right\}} \][/tex]