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Sagot :
Certainly! Let's embark on solving the system of equations step-by-step:
Given the system of equations:
[tex]\[ \begin{cases} x^2 + xy = 28 \\ y^2 + xy = 21 \end{cases} \][/tex]
We need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and then determine [tex]\( xy \)[/tex].
### Step 1: Solving the system of equations
1. Equation (1):
[tex]\[ x^2 + xy = 28 \][/tex]
2. Equation (2):
[tex]\[ y^2 + xy = 21 \][/tex]
### Step 2: Subtract Equation (2) from Equation (1) to eliminate [tex]\( xy \)[/tex]
[tex]\[ x^2 + xy - (y^2 + xy) = 28 - 21 \][/tex]
This simplifies to:
[tex]\[ x^2 - y^2 = 7 \][/tex]
Using the difference of squares, this can be written as:
[tex]\[ (x + y)(x - y) = 7 \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
We will solve [tex]\( x + y \)[/tex] and [tex]\( x - y \)[/tex].
First, let's assume different scenarios and solve accordingly:
#### Assuming [tex]\( x + y = a \)[/tex] and [tex]\( x - y = b \)[/tex]:
Then:
[tex]\[ a \cdot b = 7 \][/tex]
Substituting respective values identified (as the real solutions should satisfy initial equations):
1. If [tex]\( a = 7 \)[/tex] and [tex]\( b = 1 \)[/tex]:
- Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] :
[tex]\[ x + y = 7 \][/tex]
[tex]\[ x - y = 1 \][/tex]
Adding these equations gives:
[tex]\[ 2x = 8 \implies x = 4 \][/tex]
Subtracting them gives:
[tex]\[ 2y = 6 \implies y = 3 \][/tex]
2. If [tex]\( a = -7 \)[/tex] and [tex]\( b = -1 \)[/tex]:
- Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x + y = -7 \][/tex]
[tex]\[ x - y = -1 \][/tex]
Adding these equations gives:
[tex]\[ 2x = -8 \implies x = -4 \][/tex]
Subtracting them gives:
[tex]\[ 2y = -6 \implies y = -3 \][/tex]
Thus, the solutions to the system of equations are:
[tex]\[ x = 4, y = 3 \quad \text{or} \quad x = -4, y = -3 \][/tex]
### Step 4: Calculate [tex]\( xy \)[/tex]
For both solutions, we calculate [tex]\( xy \)[/tex]:
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ -4 \times -3 = 12 \][/tex]
Thus, for both cases:
[tex]\[ xy = 12 \][/tex]
The solutions to the system of equations are [tex]\((4, 3)\)[/tex] and [tex]\((-4, -3)\)[/tex], and in both cases, the product [tex]\( xy \)[/tex] is [tex]\( \boxed{12} \)[/tex].
Given the system of equations:
[tex]\[ \begin{cases} x^2 + xy = 28 \\ y^2 + xy = 21 \end{cases} \][/tex]
We need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] and then determine [tex]\( xy \)[/tex].
### Step 1: Solving the system of equations
1. Equation (1):
[tex]\[ x^2 + xy = 28 \][/tex]
2. Equation (2):
[tex]\[ y^2 + xy = 21 \][/tex]
### Step 2: Subtract Equation (2) from Equation (1) to eliminate [tex]\( xy \)[/tex]
[tex]\[ x^2 + xy - (y^2 + xy) = 28 - 21 \][/tex]
This simplifies to:
[tex]\[ x^2 - y^2 = 7 \][/tex]
Using the difference of squares, this can be written as:
[tex]\[ (x + y)(x - y) = 7 \][/tex]
### Step 3: Solve for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]
We will solve [tex]\( x + y \)[/tex] and [tex]\( x - y \)[/tex].
First, let's assume different scenarios and solve accordingly:
#### Assuming [tex]\( x + y = a \)[/tex] and [tex]\( x - y = b \)[/tex]:
Then:
[tex]\[ a \cdot b = 7 \][/tex]
Substituting respective values identified (as the real solutions should satisfy initial equations):
1. If [tex]\( a = 7 \)[/tex] and [tex]\( b = 1 \)[/tex]:
- Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex] :
[tex]\[ x + y = 7 \][/tex]
[tex]\[ x - y = 1 \][/tex]
Adding these equations gives:
[tex]\[ 2x = 8 \implies x = 4 \][/tex]
Subtracting them gives:
[tex]\[ 2y = 6 \implies y = 3 \][/tex]
2. If [tex]\( a = -7 \)[/tex] and [tex]\( b = -1 \)[/tex]:
- Solving for [tex]\( x \)[/tex] and [tex]\( y \)[/tex]:
[tex]\[ x + y = -7 \][/tex]
[tex]\[ x - y = -1 \][/tex]
Adding these equations gives:
[tex]\[ 2x = -8 \implies x = -4 \][/tex]
Subtracting them gives:
[tex]\[ 2y = -6 \implies y = -3 \][/tex]
Thus, the solutions to the system of equations are:
[tex]\[ x = 4, y = 3 \quad \text{or} \quad x = -4, y = -3 \][/tex]
### Step 4: Calculate [tex]\( xy \)[/tex]
For both solutions, we calculate [tex]\( xy \)[/tex]:
[tex]\[ 4 \times 3 = 12 \][/tex]
[tex]\[ -4 \times -3 = 12 \][/tex]
Thus, for both cases:
[tex]\[ xy = 12 \][/tex]
The solutions to the system of equations are [tex]\((4, 3)\)[/tex] and [tex]\((-4, -3)\)[/tex], and in both cases, the product [tex]\( xy \)[/tex] is [tex]\( \boxed{12} \)[/tex].
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