# Simultaneous Equations (Algebra) – 2017 GCSE Higher-tier Maths question (solved by a 13-year-old) Welcome to Mathematics Tube 🙂 I’m going to try a GCSE higher tier
maths question, so the question is solve algebraically the simultaneous equations.
To solve simultaneous equations, we can use two techniques which are
method of elimination and method of substitution. So, in this question, one of
the equations is a quadratic equation. So I will use the method of substitution.
First, I need to label these given equations, so, x squared plus y squared
equals to 25 will be labeled as 1 and y minus 3x equals to 13 will be labeled as 2.
Ok, so now I need to make y the subject in equation 2, so, that would be y
equals to 13 plus 3x. Now I’m going to substitute y in equation 1 with 13 plus
3x, so I will get x squared plus 13 plus 3 x the whole square equals to 25. Okay
now 13 plus 2x squared is 13 squared plus 2 times 13 times 3x plus 3x squared
equals to 25. Now I’m going to simplify this, so I will get x squared plus 169
plus 78 x plus 9 x squared equal to 25. Okay then if I further simplify this
I would get 10x squared plus 78 x + 169 minus 25 equals to 0.
Finally I will get 10 x squared plus 78 x plus 144 equals to 0. So now I got a
quadratic equation in variable x. So I’m going to solve this equation by
factorizing it. So first step I need to divide 78 into two numbers and if I
multiply those two numbers I would get the product of the constant and the
coefficient of x squared which would be 1440 but if I plus those two numbers
together I would get the coefficient of x which is 78 and those two numbers
would be 30 and 48. Okay so now I’ll just write the equation where 78
divided into two numbers, so 10 x squared plus 30x plus 48 X plus 144 equal to 0
okay next I’m just going to draw a line in
the middle because I need to factorize 10 x squared plus 30x and 48 X plus 144
separately. So, if I factorize 10 x squared plus 30x, I would get 10x
times x plus 3, the other one 48 x plus 144, I would get 48 times x plus 3 as well. So,
now, so the common bracket is x plus 3 and the numbers on the outside is 10 x +
48 so 10 x plus 48 would be 1 bracket and x plus 3 would be 1 bucket equals to
0. So if x plus 3 times 10 x plus 48 equals to 0, x plus 3 might be equal to 0
or 10x plus 48 might be equal to 0, all right, so now I can find a value of x. So
x equals to minus 3 or x equals to 48 minus 48 over 10 which is minus 24 over
5. Okay now I’m going to substitute the value of x in equation two to find the
value of y. So equation 2 is y minus 3x equals to13 and I will substitute x with minus 24 over 5. Okay
so y minus 3 times minus 24 over 5 equal to 13, so I will get y plus 72 over 5
equals to 13. So now I’m going to subtract both sides
of the equation by 72 over five, so I will get y equal to thirteen minus 72
over five or 13 over one, okay so I’m going to make the denominators the
same so 13 times 5 is 65 over five minus 72 over five, okay, okay now 65 over five
minus 72 over five is minus 7 over 5. so y would be equal to minus seven over
five, okay so to find the other one, I need to write y minus three x equals to
thirteen again, so this time I’m going to substitute x with minus three, so y minus
three times minus three equals 13, which I will get y plus 9 equals to 13, finally I
will get y equals to 13 minus 9 which is y equals to four, okay so the final
answer would be x equals to minus 3 and y equals to 4 or x equals to minus 24 over 5 and y equals to minus 7 over 5. So that is the
final answer for this question. Bye ! Please like, share and subscribe. Thank you so much for watching ! 🙂

### One thought on “Simultaneous Equations (Algebra) – 2017 GCSE Higher-tier Maths question (solved by a 13-year-old)”

• November 9, 2019 at 12:31 pm