Math Problem Statement

Simultaneous equations: x - 9y = 10 and 3y^2 = 4x + 7. Part (a): Show that 3y^2 - 36y - 47 = 0. Part (b): Solve the simultaneous equations.

Solution

Let's break down and solve the problem shown in the image.

Given Simultaneous Equations:

  1. x9y=10x - 9y = 10
  2. 3y2=4x+73y^2 = 4x + 7

Part (a) - Show that 3y236y47=03y^2 - 36y - 47 = 0

We need to manipulate the two given equations to reach this form.

Start with the first equation: x=9y+10x = 9y + 10 Now, substitute this expression for xx into the second equation: 3y2=4x+73y^2 = 4x + 7 Substitute x=9y+10x = 9y + 10 into the equation: 3y2=4(9y+10)+73y^2 = 4(9y + 10) + 7 Expand the right-hand side: 3y2=36y+40+73y^2 = 36y + 40 + 7 Simplify: 3y2=36y+473y^2 = 36y + 47 Now, subtract 36y+4736y + 47 from both sides to get: 3y236y47=03y^2 - 36y - 47 = 0 Thus, part (a) is verified.

Part (b) - Solve the Simultaneous Equations

Now, we solve the quadratic equation 3y236y47=03y^2 - 36y - 47 = 0.

We'll use the quadratic formula: y=b±b24ac2ay = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} For the equation 3y236y47=03y^2 - 36y - 47 = 0, the coefficients are:

  • a=3a = 3
  • b=36b = -36
  • c=47c = -47

Now, calculating the discriminant: Δ=(36)24(3)(47)\Delta = (-36)^2 - 4(3)(-47) Δ=1296+564=1860\Delta = 1296 + 564 = 1860 Thus: y=36±18606y = \frac{36 \pm \sqrt{1860}}{6} Approximating 1860\sqrt{1860}: 186043.13\sqrt{1860} \approx 43.13 So the two possible values of yy are: y1=36+43.13679.13613.19y_1 = \frac{36 + 43.13}{6} \approx \frac{79.13}{6} \approx 13.19 y2=3643.1367.1361.19y_2 = \frac{36 - 43.13}{6} \approx \frac{-7.13}{6} \approx -1.19

Now, substitute these values of yy back into the equation x=9y+10x = 9y + 10 to find the corresponding values of xx.

For y1=13.19y_1 = 13.19: x1=9(13.19)+10118.71+10=128.71x_1 = 9(13.19) + 10 \approx 118.71 + 10 = 128.71

For y2=1.19y_2 = -1.19: x2=9(1.19)+1010.71+10=0.71x_2 = 9(-1.19) + 10 \approx -10.71 + 10 = -0.71

Thus, the solutions are approximately: (x1,y1)(128.7,13.2)(x_1, y_1) \approx (128.7, 13.2) (x2,y2)(0.7,1.2)(x_2, y_2) \approx (-0.7, -1.2)

Would you like any further details or clarifications on these steps?

Related Questions:

  1. What are the key steps for solving quadratic equations using the quadratic formula?
  2. How would the method change if the simultaneous equations involved more variables?
  3. What happens if the discriminant in the quadratic formula is negative?
  4. Can we solve these equations using substitution and elimination without the quadratic formula?
  5. How can this method be applied to real-world problems involving simultaneous equations?

Tip:

When solving simultaneous equations, always simplify one equation to express one variable in terms of the other before substituting into the second equation. This reduces complexity.

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Math Problem Analysis

Mathematical Concepts

Simultaneous Equations
Quadratic Equations
Algebra

Formulas

Quadratic Formula: y = (-b ± sqrt(b^2 - 4ac)) / 2a

Theorems

Quadratic Formula
Substitution Method

Suitable Grade Level

Grades 10-12