Discover the answers you need at Westonci.ca, where experts provide clear and concise information on various topics. Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
To solve the system of simultaneous equations:
[tex]\[ \begin{aligned} x & = 7 - 3y \\ x^2 - y^2 & = 39 \end{aligned} \][/tex]
we'll proceed step-by-step:
### Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation
Given:
[tex]\[ x = 7 - 3y \][/tex]
Substitute [tex]\( x = 7 - 3y \)[/tex] into the second equation [tex]\( x^2 - y^2 = 39 \)[/tex]:
[tex]\[ (7 - 3y)^2 - y^2 = 39 \][/tex]
### Step 2: Expand and simplify
Expand [tex]\( (7 - 3y)^2 \)[/tex]:
[tex]\[ (7 - 3y)^2 = 49 - 42y + 9y^2 \][/tex]
Now substitute this back in:
[tex]\[ 49 - 42y + 9y^2 - y^2 = 39 \][/tex]
Simplify the equation:
[tex]\[ 49 - 42y + 8y^2 = 39 \][/tex]
### Step 3: Combine like terms and solve for [tex]\( y \)[/tex]:
[tex]\[ 8y^2 - 42y + 49 - 39 = 0 \][/tex]
[tex]\[ 8y^2 - 42y + 10 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve [tex]\( 8y^2 - 42y + 10 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -42 \)[/tex], and [tex]\( c = 10 \)[/tex].
So,
[tex]\[ y = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 8 \cdot 10}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1764 - 320}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1444}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm 38}{16} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{42 + 38}{16} = \frac{80}{16} = 5 \][/tex]
or
[tex]\[ y = \frac{42 - 38}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
### Step 5: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]:
1. When [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 7 - 3 \cdot 5 = 7 - 15 = -8 \][/tex]
2. When [tex]\( y = \frac{1}{4} \)[/tex]:
[tex]\[ x = 7 - 3 \cdot \frac{1}{4} = 7 - \frac{3}{4} = \frac{28}{4} - \frac{3}{4} = \frac{25}{4} \][/tex]
### Step 6: List the solutions
The solutions for the system of equations are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
Thus, the correct solutions are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
[tex]\[ \begin{aligned} x & = 7 - 3y \\ x^2 - y^2 & = 39 \end{aligned} \][/tex]
we'll proceed step-by-step:
### Step 1: Substitute [tex]\( x \)[/tex] from the first equation into the second equation
Given:
[tex]\[ x = 7 - 3y \][/tex]
Substitute [tex]\( x = 7 - 3y \)[/tex] into the second equation [tex]\( x^2 - y^2 = 39 \)[/tex]:
[tex]\[ (7 - 3y)^2 - y^2 = 39 \][/tex]
### Step 2: Expand and simplify
Expand [tex]\( (7 - 3y)^2 \)[/tex]:
[tex]\[ (7 - 3y)^2 = 49 - 42y + 9y^2 \][/tex]
Now substitute this back in:
[tex]\[ 49 - 42y + 9y^2 - y^2 = 39 \][/tex]
Simplify the equation:
[tex]\[ 49 - 42y + 8y^2 = 39 \][/tex]
### Step 3: Combine like terms and solve for [tex]\( y \)[/tex]:
[tex]\[ 8y^2 - 42y + 49 - 39 = 0 \][/tex]
[tex]\[ 8y^2 - 42y + 10 = 0 \][/tex]
### Step 4: Solve the quadratic equation
To solve [tex]\( 8y^2 - 42y + 10 = 0 \)[/tex], we can use the quadratic formula:
[tex]\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Here, [tex]\( a = 8 \)[/tex], [tex]\( b = -42 \)[/tex], and [tex]\( c = 10 \)[/tex].
So,
[tex]\[ y = \frac{42 \pm \sqrt{(-42)^2 - 4 \cdot 8 \cdot 10}}{2 \cdot 8} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1764 - 320}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm \sqrt{1444}}{16} \][/tex]
[tex]\[ y = \frac{42 \pm 38}{16} \][/tex]
This gives us two solutions for [tex]\( y \)[/tex]:
[tex]\[ y = \frac{42 + 38}{16} = \frac{80}{16} = 5 \][/tex]
or
[tex]\[ y = \frac{42 - 38}{16} = \frac{4}{16} = \frac{1}{4} \][/tex]
### Step 5: Substitute [tex]\( y \)[/tex] back to find [tex]\( x \)[/tex]:
1. When [tex]\( y = 5 \)[/tex]:
[tex]\[ x = 7 - 3 \cdot 5 = 7 - 15 = -8 \][/tex]
2. When [tex]\( y = \frac{1}{4} \)[/tex]:
[tex]\[ x = 7 - 3 \cdot \frac{1}{4} = 7 - \frac{3}{4} = \frac{28}{4} - \frac{3}{4} = \frac{25}{4} \][/tex]
### Step 6: List the solutions
The solutions for the system of equations are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
Thus, the correct solutions are:
[tex]\[ (x, y) = (-8, 5) \quad \text{and} \quad \left( \frac{25}{4}, \frac{1}{4} \right) \][/tex]
Thanks for using our service. We aim to provide the most accurate answers for all your queries. Visit us again for more insights. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Thank you for trusting Westonci.ca. Don't forget to revisit us for more accurate and insightful answers.