MYTUTOR SUBJECT ANSWERS

173 views

Solve the Simultaneous Equations -3X + 4Y=11 & X-2Y = -5 to find the values of X and Y

Simultaneous Equations are equations involving 2 or more unknowns that have the same value in each equation. The aim is to eliminate all but one unknown. This will then allow us to find the value of the non eliminated unknown. From this any other unknowns can be found. There are 2 main methods we can use to solve these problems:

A) Solving by Elimination.

The first step is to label the equations

(1) -3X +4y= 11

(2)    X - 2Y= -5

You must then manipulate the equations in such a way that the coefficient of one of the unknown variables is the same value in both equations. 

In this example, the equations will be manipulated in order to obtain equal values for the coefficient of Y in both equations

Equation (2) multiplied by 2

(1):        -3X +4y= 11

(2) x 2:   2X -4Y= -10

We can now add both equations together in order to eliminate the Y unknown.

(1) + (2) x 2

(1):        -3X +4y= 11

              +

(2) x 2:   2X -4Y= -10

-X= 1

Divide both sides by -1

X= -1

In order to find the value of Y, substitute X= -1 into any one of the original equations.

Substituting X= -1 into  equation (2):      X- 2Y = -5

  

-1 -2Y = -5

-2Y = -4

Divide both sides by -2

Y=2

Therefore we have solved the simultaneous equations and obtained the answer of X= -1 and Y=2.

B) Solving by Substitution

The first step here is making one of the unknown variables the subject of one of the equations.

Using equation (2): X- 2Y= -5 ——> X= -5 -2Y

You must then substitute this value for the unknown into the other equation.

Substituting X= -5- 2Y into Equation (1): -3X + 4Y =11

-3 (-5+ 2Y) +4Y =11

Expand the brackets

15 -6Y + 4Y =11

Group like terms together

15-11 = 6Y-4Y

4= 2Y

Divide both sides by 2

Y=2

To find the value of X, substitute  Y=2 into X= -5- 2Y

X= -5 - 2(2)

X= -1

Therefore we have solved the simultaneous equations and obtained the answer of x= -1 and y=2.

Emehakon U. GCSE Maths tutor, GCSE Law tutor, A Level Law tutor

5 months ago

Answered by Emehakon, a GCSE Maths tutor with MyTutor

Still stuck? Get one-to-one help from a personally interviewed subject specialist

331 SUBJECT SPECIALISTS

£18 /hr

Emma W.

Degree: Natural Sciences (Bachelors) - Durham University

Subjects offered: Maths, Physics+ 2 more

Maths
Physics
Chemistry
Biology

“Hi! I'm Emma and I'm a second year Natural Sciences student from Durham University, which means I've studied biology, physics and maths at university level. I have a very friendly tutoring style as I believe students are most able to ...”

£18 /hr

Sam W.

Degree: Mathematics (Bachelors) - Durham University

Subjects offered: Maths

Maths

“About me Hi! I'm Sam and I'm a second year mathematics student at Durham University.  In terms of my academic background, I achieved an A* at GCSE maths, before going on to study maths and further maths at A level and obtaining a high...”

MyTutor guarantee

£18 /hr

Wesley M.

Degree: Mathematics (Masters) - Bristol University

Subjects offered: Maths, Physics+ 1 more

Maths
Physics
Chemistry

“I am a current Mathematics student at The University of Bristol. Maths can be a love-hate sort of subject for many people, but with my absolute love and enthusiasm for both maths itself and the idea of inspiring others, I hope to be a...”

About the author

£18 /hr

Emehakon U.

Degree: Law (Bachelors) - Durham University

Subjects offered: Maths, Law

Maths
Law

“I am a Law student at Durham University. I have always had a passion for finding solutions to problems. (Both in a practical and Mathematical sense!) . I'll be sure to utilize that passion in order to give you the best outcome possibl...”

MyTutor guarantee

You may also like...

Other GCSE Maths questions

Factorise x^2 + 7x + 10

Solve these simultaneous equations. 2x + y = 10 and 3x + 4y = 25.

Finding Roots of Quadratic Equations

Find the mean, median, mode and range of this data: 2, 5,6,12,5

View GCSE Maths tutors

Cookies:

We use cookies to improve our service. By continuing to use this website, we'll assume that you're OK with this. Dismiss

mtw:mercury1:status:ok