Answer :
To solve the system of simultaneous equations given by:
1. \( x^2 + y^2 = 13 \)
2. \( x = y - 5 \)
we will follow these steps:
Step 1: Substitute the expression for \( x \) from the second equation into the first equation.
Given \( x = y - 5 \), substitute \( x \) in the first equation:
[tex]\[ (y - 5)^2 + y^2 = 13 \][/tex]
Step 2: Expand and simplify the equation.
First, expand \( (y - 5)^2 \):
[tex]\[ (y - 5)^2 = y^2 - 10y + 25 \][/tex]
So the equation becomes:
[tex]\[ y^2 - 10y + 25 + y^2 = 13 \][/tex]
Combine like terms:
[tex]\[ 2y^2 - 10y + 25 = 13 \][/tex]
Step 3: Rearrange the equation into standard quadratic form.
[tex]\[ 2y^2 - 10y + 25 - 13 = 0 \][/tex]
[tex]\[ 2y^2 - 10y + 12 = 0 \][/tex]
Step 4: Simplify the quadratic equation.
Divide every term by 2 to make it easier to solve:
[tex]\[ y^2 - 5y + 6 = 0 \][/tex]
Step 5: Factor the quadratic equation.
[tex]\[ y^2 - 5y + 6 = (y - 2)(y - 3) = 0 \][/tex]
Set each factor to zero to solve for \( y \):
[tex]\[ y - 2 = 0 \quad \text{or} \quad y - 3 = 0 \][/tex]
So,
[tex]\[ y = 2 \quad \text{or} \quad y = 3 \][/tex]
Step 6: Find the corresponding \( x \) values.
Using the second original equation \( x = y - 5 \):
- When \( y = 2 \):
[tex]\[ x = 2 - 5 = -3 \][/tex]
- When \( y = 3 \):
[tex]\[ x = 3 - 5 = -2 \][/tex]
Step 7: Write the solutions as pairs \((x, y)\).
The solutions to the system of equations are:
[tex]\[ (-3, 2) \][/tex]
[tex]\[ (-2, 3) \][/tex]
Thus, the pairs [tex]\((x, y)\)[/tex] that satisfy both equations are [tex]\((-3, 2)\)[/tex] and [tex]\((-2, 3)\)[/tex].
1. \( x^2 + y^2 = 13 \)
2. \( x = y - 5 \)
we will follow these steps:
Step 1: Substitute the expression for \( x \) from the second equation into the first equation.
Given \( x = y - 5 \), substitute \( x \) in the first equation:
[tex]\[ (y - 5)^2 + y^2 = 13 \][/tex]
Step 2: Expand and simplify the equation.
First, expand \( (y - 5)^2 \):
[tex]\[ (y - 5)^2 = y^2 - 10y + 25 \][/tex]
So the equation becomes:
[tex]\[ y^2 - 10y + 25 + y^2 = 13 \][/tex]
Combine like terms:
[tex]\[ 2y^2 - 10y + 25 = 13 \][/tex]
Step 3: Rearrange the equation into standard quadratic form.
[tex]\[ 2y^2 - 10y + 25 - 13 = 0 \][/tex]
[tex]\[ 2y^2 - 10y + 12 = 0 \][/tex]
Step 4: Simplify the quadratic equation.
Divide every term by 2 to make it easier to solve:
[tex]\[ y^2 - 5y + 6 = 0 \][/tex]
Step 5: Factor the quadratic equation.
[tex]\[ y^2 - 5y + 6 = (y - 2)(y - 3) = 0 \][/tex]
Set each factor to zero to solve for \( y \):
[tex]\[ y - 2 = 0 \quad \text{or} \quad y - 3 = 0 \][/tex]
So,
[tex]\[ y = 2 \quad \text{or} \quad y = 3 \][/tex]
Step 6: Find the corresponding \( x \) values.
Using the second original equation \( x = y - 5 \):
- When \( y = 2 \):
[tex]\[ x = 2 - 5 = -3 \][/tex]
- When \( y = 3 \):
[tex]\[ x = 3 - 5 = -2 \][/tex]
Step 7: Write the solutions as pairs \((x, y)\).
The solutions to the system of equations are:
[tex]\[ (-3, 2) \][/tex]
[tex]\[ (-2, 3) \][/tex]
Thus, the pairs [tex]\((x, y)\)[/tex] that satisfy both equations are [tex]\((-3, 2)\)[/tex] and [tex]\((-2, 3)\)[/tex].
