Answer :
To solve the system of equations, we need to find the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] that satisfy all three of the equations simultaneously. The system given is:
1. [tex]\( y = 2x^2 + 3 \)[/tex]
2. [tex]\( 5x^2 - 2y = x \)[/tex]
3. [tex]\( 3y - 4x = 51 \)[/tex]
Let's solve this step-by-step.
### Step 1: Substitute [tex]\( y \)[/tex] from Equation 1 into Equations 2 and 3
Substitute [tex]\( y = 2x^2 + 3 \)[/tex] into the second equation [tex]\( 5x^2 - 2y = x \)[/tex]:
[tex]\[ 5x^2 - 2(2x^2 + 3) = x \][/tex]
[tex]\[ 5x^2 - 4x^2 - 6 = x \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]
This is a quadratic equation. Let's solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} \][/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]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ x = \frac{1 - 5}{2} = -2 \][/tex]
### Step 2: Find Corresponding [tex]\( y \)[/tex] Values
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3)^2 + 3 \][/tex]
[tex]\[ y = 2 \times 9 + 3 \][/tex]
[tex]\[ y = 18 + 3 \][/tex]
[tex]\[ y = 21 \][/tex]
For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 2(-2)^2 + 3 \][/tex]
[tex]\[ y = 2 \times 4 + 3 \][/tex]
[tex]\[ y = 8 + 3 \][/tex]
[tex]\[ y = 11 \][/tex]
### Step 3: Verify in the Third Equation
Check if both pairs [tex]\((x, y) = (3, 21)\)[/tex] and [tex]\((-2, 11)\)[/tex] satisfy the third equation [tex]\( 3y - 4x = 51 \)[/tex]:
For [tex]\( (3, 21) \)[/tex]:
[tex]\[ 3(21) - 4(3) = 51 \][/tex]
[tex]\[ 63 - 12 = 51 \][/tex]
[tex]\[ 51 = 51 \][/tex]
This pair is a solution.
For [tex]\( (-2, 11) \)[/tex]:
[tex]\[ 3(11) - 4(-2) = 51 \][/tex]
[tex]\[ 33 + 8 = 41 \][/tex]
[tex]\[ 41 \neq 51 \][/tex]
This pair is not a solution.
### Conclusion
The solution to the system of equations is:
[tex]\[ x = 3 \][/tex]
[tex]\[ y = 21 \][/tex]
Therefore, the only solution to the system is [tex]\((3, 21)\)[/tex].
1. [tex]\( y = 2x^2 + 3 \)[/tex]
2. [tex]\( 5x^2 - 2y = x \)[/tex]
3. [tex]\( 3y - 4x = 51 \)[/tex]
Let's solve this step-by-step.
### Step 1: Substitute [tex]\( y \)[/tex] from Equation 1 into Equations 2 and 3
Substitute [tex]\( y = 2x^2 + 3 \)[/tex] into the second equation [tex]\( 5x^2 - 2y = x \)[/tex]:
[tex]\[ 5x^2 - 2(2x^2 + 3) = x \][/tex]
[tex]\[ 5x^2 - 4x^2 - 6 = x \][/tex]
[tex]\[ x^2 - x - 6 = 0 \][/tex]
This is a quadratic equation. Let's solve for [tex]\( x \)[/tex] using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -1 \)[/tex], and [tex]\( c = -6 \)[/tex]:
[tex]\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-6)}}{2(1)} \][/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]
So, the solutions for [tex]\( x \)[/tex] are:
[tex]\[ x = \frac{1 + 5}{2} = 3 \][/tex]
[tex]\[ x = \frac{1 - 5}{2} = -2 \][/tex]
### Step 2: Find Corresponding [tex]\( y \)[/tex] Values
For [tex]\( x = 3 \)[/tex]:
[tex]\[ y = 2(3)^2 + 3 \][/tex]
[tex]\[ y = 2 \times 9 + 3 \][/tex]
[tex]\[ y = 18 + 3 \][/tex]
[tex]\[ y = 21 \][/tex]
For [tex]\( x = -2 \)[/tex]:
[tex]\[ y = 2(-2)^2 + 3 \][/tex]
[tex]\[ y = 2 \times 4 + 3 \][/tex]
[tex]\[ y = 8 + 3 \][/tex]
[tex]\[ y = 11 \][/tex]
### Step 3: Verify in the Third Equation
Check if both pairs [tex]\((x, y) = (3, 21)\)[/tex] and [tex]\((-2, 11)\)[/tex] satisfy the third equation [tex]\( 3y - 4x = 51 \)[/tex]:
For [tex]\( (3, 21) \)[/tex]:
[tex]\[ 3(21) - 4(3) = 51 \][/tex]
[tex]\[ 63 - 12 = 51 \][/tex]
[tex]\[ 51 = 51 \][/tex]
This pair is a solution.
For [tex]\( (-2, 11) \)[/tex]:
[tex]\[ 3(11) - 4(-2) = 51 \][/tex]
[tex]\[ 33 + 8 = 41 \][/tex]
[tex]\[ 41 \neq 51 \][/tex]
This pair is not a solution.
### Conclusion
The solution to the system of equations is:
[tex]\[ x = 3 \][/tex]
[tex]\[ y = 21 \][/tex]
Therefore, the only solution to the system is [tex]\((3, 21)\)[/tex].
