Solve algebraically the simultaneous equations
x^{2} + y^{2} = 25
y - 3x = 13
(Total for Question 20 is 5 marks) - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1

Question 20

Solve algebraically the simultaneous equations
x^{2} + y^{2} = 25
y - 3x = 13
(Total for Question 20 is 5 marks)
Worked Solution & Example Answer:Solve algebraically the simultaneous equations
x^{2} + y^{2} = 25
y - 3x = 13
(Total for Question 20 is 5 marks) - Edexcel - GCSE Maths - Question 20 - 2017 - Paper 1
Substituting for $y$

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From the second equation, rearrange to find y:
y=3x+13.
Now substitute this expression for y into the first equation.
Expanding the first equation

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Substituting gives us:
x2+(3x+13)2=25.
Expanding (3x+13)2 yields:
x2+(9x2+78x+169)=25.
Combining like terms

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Combine the terms:
10x2+78x+169=25.
Subtract 25 from both sides to set the equation to zero:
10x2+78x+144=0.
Factoring

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Now, factor the quadratic equation, if possible:
(5x + 12)(2x + 12) = 0.$$Finding the values of $x$

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Setting each factor to zero gives:
2x + 12 = 0.$$
This leads to $x = -rac{12}{5}$ and $x = -6$.Finding the corresponding $y$ values

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Substituting back into y=3x+13:
- 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:
(-rac{12}{5}, rac{29}{5}) and (−6,−5).
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