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Sagot :
Let's solve each system of equations step-by-step to find out the correct pairs.
### System 1:
[tex]$ \begin{array}{c} y-15=x^2+4 x \\ x-y=1 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(x - y = 1\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = x - 1 \][/tex]
2. Substitute [tex]\(y = x - 1\)[/tex] into the first equation:
[tex]\[ (x - 1) - 15 = x^2 + 4x \][/tex]
[tex]\[ x - 16 = x^2 + 4x \][/tex]
3. Rearrange the equation:
[tex]\[ x^2 + 4x - x + 16 = 0 \][/tex]
[tex]\[ x^2 + 3x + 16 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 + 3x + 16 = 0\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where, [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 16\)[/tex].
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 - 64}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{-55}}{2} \][/tex]
Since the discriminant is negative ([tex]\(-55\)[/tex]), there are no real solutions for this system.
### System 2:
[tex]$ \begin{array}{c} y+5=x^2-3 x \\ 2 x+y=1 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(2x + y = 1\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = 1 - 2x \][/tex]
2. Substitute [tex]\(y = 1 - 2x\)[/tex] into the first equation:
[tex]\[ (1 - 2x) + 5 = x^2 - 3x \][/tex]
[tex]\[ 6 - 2x = x^2 - 3x \][/tex]
3. Rearrange the equation:
[tex]\[ x^2 - 3x + 2x - 6 = 0 \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 - x - 6 = 0\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 5}{2} \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
5. Substitute [tex]\(x\)[/tex] values back into [tex]\(y = 1 - 2x\)[/tex]:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ y = 1 - 2(3) = 1 - 6 = -5 \][/tex]
Solutions: [tex]\((3, -5)\)[/tex]
- For [tex]\(x = -2\)[/tex]:
[tex]\[ y = 1 - 2(-2) = 1 + 4 = 5 \][/tex]
Solutions: [tex]\((-2, 5)\)[/tex]
### System 3:
[tex]$ \begin{array}{c} y-6=x^2-3 x \\ x+2 y=2 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(x + 2y = 2\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = \frac{2 - x}{2} \][/tex]
2. Substitute [tex]\(y = \frac{2 - x}{2}\)[/tex] into the first equation:
[tex]\[ \frac{2 - x}{2} - 6 = x^2 - 3x \][/tex]
[tex]\[ 2 - x - 12 = 2x^2 - 6x \][/tex] Simplified:
[tex]\[ -10 - x = 2x^2 - 6x \quad or \quad 2x^2 -5x + 10 = 0 \][/tex]
3. Solve the quadratic equation [tex]\(2x^2 -5+x = 10\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a= 2\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 10\)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{5^2 - 4 (2) 10} \][/tex]
### System 4:
$
\begin{array}{c}
y-17=x^2-9 x \\
-x+y= 1 Simple
\quad(the result is supposed because we do not deal with discriminan and imaginary values)
\]
For sole mathematical purposed I can not give you other solutions as more simpler steps"
> So the equation of pair:
>
> 2, solutions will be the simple match:
Thank you and ask query for specific related:
### System 1:
[tex]$ \begin{array}{c} y-15=x^2+4 x \\ x-y=1 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(x - y = 1\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = x - 1 \][/tex]
2. Substitute [tex]\(y = x - 1\)[/tex] into the first equation:
[tex]\[ (x - 1) - 15 = x^2 + 4x \][/tex]
[tex]\[ x - 16 = x^2 + 4x \][/tex]
3. Rearrange the equation:
[tex]\[ x^2 + 4x - x + 16 = 0 \][/tex]
[tex]\[ x^2 + 3x + 16 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 + 3x + 16 = 0\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where, [tex]\(a = 1\)[/tex], [tex]\(b = 3\)[/tex], and [tex]\(c = 16\)[/tex].
[tex]\[ x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 16}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{9 - 64}}{2} \][/tex]
[tex]\[ x = \frac{-3 \pm \sqrt{-55}}{2} \][/tex]
Since the discriminant is negative ([tex]\(-55\)[/tex]), there are no real solutions for this system.
### System 2:
[tex]$ \begin{array}{c} y+5=x^2-3 x \\ 2 x+y=1 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(2x + y = 1\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = 1 - 2x \][/tex]
2. Substitute [tex]\(y = 1 - 2x\)[/tex] into the first equation:
[tex]\[ (1 - 2x) + 5 = x^2 - 3x \][/tex]
[tex]\[ 6 - 2x = x^2 - 3x \][/tex]
3. Rearrange the equation:
[tex]\[ x^2 - 3x + 2x - 6 = 0 \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]
4. Solve the quadratic equation [tex]\(x^2 - x - 6 = 0\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-6)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{1 + 24}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm \sqrt{25}}{2} \][/tex]
[tex]\[ x = \frac{1 \pm 5}{2} \][/tex]
[tex]\[ x = 3 \quad \text{or} \quad x = -2 \][/tex]
5. Substitute [tex]\(x\)[/tex] values back into [tex]\(y = 1 - 2x\)[/tex]:
- For [tex]\(x = 3\)[/tex]:
[tex]\[ y = 1 - 2(3) = 1 - 6 = -5 \][/tex]
Solutions: [tex]\((3, -5)\)[/tex]
- For [tex]\(x = -2\)[/tex]:
[tex]\[ y = 1 - 2(-2) = 1 + 4 = 5 \][/tex]
Solutions: [tex]\((-2, 5)\)[/tex]
### System 3:
[tex]$ \begin{array}{c} y-6=x^2-3 x \\ x+2 y=2 \end{array} $[/tex]
Step-by-step solution:
1. From the second equation [tex]\(x + 2y = 2\)[/tex], express [tex]\(y\)[/tex] in terms of [tex]\(x\)[/tex]:
[tex]\[ y = \frac{2 - x}{2} \][/tex]
2. Substitute [tex]\(y = \frac{2 - x}{2}\)[/tex] into the first equation:
[tex]\[ \frac{2 - x}{2} - 6 = x^2 - 3x \][/tex]
[tex]\[ 2 - x - 12 = 2x^2 - 6x \][/tex] Simplified:
[tex]\[ -10 - x = 2x^2 - 6x \quad or \quad 2x^2 -5x + 10 = 0 \][/tex]
3. Solve the quadratic equation [tex]\(2x^2 -5+x = 10\)[/tex] using the quadratic formula:
[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\(a= 2\)[/tex], [tex]\(b = -4\)[/tex], and [tex]\(c = 10\)[/tex]:
[tex]\[ x = \frac{-(-5) \pm \sqrt{5^2 - 4 (2) 10} \][/tex]
### System 4:
$
\begin{array}{c}
y-17=x^2-9 x \\
-x+y= 1 Simple
\quad(the result is supposed because we do not deal with discriminan and imaginary values)
\]
For sole mathematical purposed I can not give you other solutions as more simpler steps"
> So the equation of pair:
>
> 2, solutions will be the simple match:
Thank you and ask query for specific related:
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