To solve the simultaneous equations [tex]\( y = 3x + 5 \)[/tex] and [tex]\( y = 4x^2 + x \)[/tex], we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy both equations.
Here are the detailed steps:
1. Set the equations equal to each other:
Since both equations equal [tex]\( y \)[/tex], we can set them equal to each other to find [tex]\( x \)[/tex]:
[tex]\[
3x + 5 = 4x^2 + x
\][/tex]
2. Rearrange to form a quadratic equation:
Move all terms to one side to set the equation to 0:
[tex]\[
4x^2 + x - 3x - 5 = 0
\][/tex]
Simplify the equation:
[tex]\[
4x^2 - 2x - 5 = 0
\][/tex]
3. Solve the quadratic equation:
We can solve the quadratic equation [tex]\( 4x^2 - 2x - 5 = 0 \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 4 \)[/tex], [tex]\( b = -2 \)[/tex], and [tex]\( c = -5 \)[/tex].
4. Find [tex]\( x \)[/tex] values:
Apply the quadratic formula:
[tex]\[
x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 4 \cdot (-5)}}{2 \cdot 4}
\][/tex]
Simplify inside the square root:
[tex]\[
x = \frac{2 \pm \sqrt{4 + 80}}{8}
\][/tex]
[tex]\[
x = \frac{2 \pm \sqrt{84}}{8}
\][/tex]
Simplify [tex]\(\sqrt{84}\)[/tex]:
[tex]\[
x = \frac{2 \pm 2\sqrt{21}}{8}
\][/tex]
Simplify further:
[tex]\[
x = \frac{1 \pm \sqrt{21}}{4}
\][/tex]
5. Calculate the numerical values of [tex]\( x \)[/tex]:
[tex]\[
x_1 \approx -0.90
\][/tex]
[tex]\[
x_2 \approx 1.40
\][/tex]
6. Find the corresponding [tex]\( y \)[/tex] values:
Substitute each [tex]\( x \)[/tex] back into one of the original equations to find [tex]\( y \)[/tex]. We'll use [tex]\( y = 3x + 5 \)[/tex]:
- For [tex]\( x_1 \approx -0.90 \)[/tex]:
[tex]\[
y_1 = 3(-0.90) + 5 \approx 2.31
\][/tex]
- For [tex]\( x_2 \approx 1.40 \)[/tex]:
[tex]\[
y_2 = 3(1.40) + 5 \approx 9.19
\][/tex]
Thus, the solutions to the simultaneous equations are:
[tex]\[
\begin{array}{l}
x \approx -0.90, \quad y \approx 2.31 \\
x \approx 1.40, \quad y \approx 9.19
\end{array}
\][/tex]
These values are rounded to 2 decimal places as requested.
