Solve the simultaneous equations: 5x + y = 21, x - 3y = 9

5x + y = 21x - 3y = 9First, multiply the equation one by 3 so we can cancel out the y's.15x + 3y = 63 (we've now multiplied this by 3)x - 3y = 9Now we can add the two equations together, thus adding 3 and -3, cancelling the y's out.15x + x = 16x3y + (-3y) = 0y (or in other words, just 0)63 + 9 = 72So, the added equation is:16x + 0 = 72Or, in other words16x = 72Now divide each side by 16 to get x:x = 72/16 x = 4.5Now, sub the value of x back into the first of the original equations to get y:5x + y = 21(5 x 4.5) + y = 2122.5 + y = 21 (now takeaway 22.5 from both sides)y = -1.5 To check this is correct, you can sub x = 4.5 into the second original equation to get the same value of y:x - 3y = 94.5 - 3y = 9 (now takeaway 4.5 from each side)-3y = 4.5 (now times both sides by -1 to get a positive value of y)3y = -4.5 (now divide both sides by 3)y = -1.5Both these values of y are the same so the equation checks outThe final answers are:x = 4.5 y = -1.5

Answered by Sebastian P. Maths tutor

2479 Views

See similar Maths GCSE tutors

Related Maths GCSE answers

All answers ▸

Solve 3(2x + 5) = 4 – x


Find x when: (2^x)(e^(3x+1))=10. Give your answer in the form (a + ln(b)) / (c + ln(d)) , where a,b,c,d are integers.


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


What is 25% of 400?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences