GCSE Edexcel Maths Solving Quadratic Equations: Solve algebraically the simultan

Solve algebraically the simultaneous equations $$x^2 + y^2 = 25$$ $$y - 3x = 13$$ - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

Question 20

Solve algebraically the simultaneous equations $$x^2 + y^2 = 25$$ $$y - 3x = 13$$

Worked Solution & Example Answer:Solve algebraically the simultaneous equations $$x^2 + y^2 = 25$$ $$y - 3x = 13$$ - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

Step 1

Substituting for y in terms of x

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Answer

From the second equation, we can express y in terms of x: y = 3x + 13.

Step 2

Substituting y into the first equation

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Answer

Substituting the expression for y into the first equation gives us: $x^2 + (3x + 13)^2 = 25.$

Step 3

Expanding the equation

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Answer

Expanding the left-hand side: $x^2 + (9x^2 + 78x + 169) = 25.$ This simplifies to: $10x^2 + 78x + 169 - 25 = 0,$ which further simplifies to: $10x^2 + 78x + 144 = 0.$

Step 4

Factoring the quadratic equation

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Answer

We will factor the quadratic equation: $10x^2 + 78x + 144$ . It can be factored into: $2(5x + 12)(x + 6) = 0.$ Setting each factor to zero gives: $x = - rac{12}{5} \text{ or } x = -6.$

Step 5

Finding corresponding y values

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Answer

Now substituting these x values back into the equation for y:

For $x = - rac{12}{5}$ : $y = 3(- rac{12}{5}) + 13 = - rac{36}{5} + 13 = rac{29}{5}.$
For $x = -6$ : $y = 3(-6) + 13 = -18 + 13 = -5.$ Thus, the solutions are: $(x, y) = \left(-\frac{12}{5}, \frac{29}{5}\right) \text{ and } (-6, -5).$

Solve algebraically the simultaneous equations $$x^2 + y^2 = 25$$ $$y - 3x = 13$$ - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

Worked Solution & Example Answer:Solve algebraically the simultaneous equations $$x^2 + y^2 = 25$$ $$y - 3x = 13$$ - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

Substituting for y in terms of x

Substituting y into the first equation

Expanding the equation

Factoring the quadratic equation

Finding corresponding y values

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