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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

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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.png

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

Step 1

Substituting for $y$

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Answer

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

Step 2

Expanding the first equation

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Answer

Substituting gives us: x2+(3x+13)2=25x^{2} + (3x + 13)^{2} = 25. Expanding (3x+13)2(3x + 13)^{2} yields: x2+(9x2+78x+169)=25.x^{2} + (9x^{2} + 78x + 169) = 25.

Step 3

Combining like terms

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Answer

Combine the terms: 10x2+78x+169=2510x^{2} + 78x + 169 = 25. Subtract 25 from both sides to set the equation to zero: 10x2+78x+144=0.10x^{2} + 78x + 144 = 0.

Step 4

Factoring

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Answer

Now, factor the quadratic equation, if possible:

(5x + 12)(2x + 12) = 0.$$

Step 5

Finding the values of $x$

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Answer

Setting each factor to zero gives:

2x + 12 = 0.$$ This leads to $x = - rac{12}{5}$ and $x = -6$.

Step 6

Finding the corresponding $y$ values

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Answer

Substituting back into y=3x+13y = 3x + 13:

  1. For x = - rac{12}{5}: y = 3(- rac{12}{5}) + 13 = - rac{36}{5} + 13 = rac{29}{5}.
  2. For x=6x = -6: y=3(6)+13=18+13=5.y = 3(-6) + 13 = -18 + 13 = -5. Thus the solutions are: (- rac{12}{5}, rac{29}{5}) and (6,5)(-6, -5).

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