Simultaneous Equations (Elimination)?

Question

How do I go about this simultaneous equation

4x+3y=13
x+2y=-3

I understand that I have to do something to make one of the terms the same as the other, but i'm not sure which method (x,/,LCD,HCF, etc.) that i am meant to use

Answer 1

We can use two methods, elimination and substitution.



Elimination



Multiplying the second equation by 4 we get:



4x + 8y = -12...(2)



4x + 3y = 13...(1)



(2) - (1) => 5y = -25



i.e. y = -5



Into the second equation gives:



x + 2(-5) = -3



so, x - 10 = -3



i.e. x = 7



Putting both values into the first equation to check gives:



4(7) + 3(-5) => 28 - 15 = 13...correct



Substitution



Eqaution 2 can be written as x = -2y - 3, so substituting into eqation 1 for x gives:



4(-2y - 3) + 3y = 13



so, -8y - 12 + 3y = 13



=> -5y = 25



Again, y = -5 => x = 7



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Answer 2

4x + 3y = 13

-4x - 8y = 12_____ADD



-5y = 25

y = - 5



x - 10 = - 3

x = 7



x = 7 , y= - 5

Answer 3

4x + 3y = 13 (equation 1)

x + 2y = -3 (equation 2)



multiply equation 2 by -4 then add the result to equation to eliminate x.

(x + 2y = -3)(-4) ---> -4x - 8y = 12 (add this to equation 1)



4x + 3y = 13

-4x - 8y = 12

____________

-5y = 25

y = -5



now solve for x using any of the given equations by substituting -5 to y. x must be 7

Answer 4

Your question: "Simultaneous Equations (Elimination)? How do I go about this simultaneous equation

4x+3y=13

x+2y=-3

I understand that I have to do something to make one of the terms the same as the other, but i'm not sure which method (x,/,LCD,HCF, etc.) that i am meant to use"



According to my calculator x = 7, y = -5.



There necessarily is a solution because the two lines are not parallel, so they must intersect at some point.



The coordinates of that point are the solution of the equation.



At this time I am unable to calculate correctly and have little light to work in, too.



There are other technical difficulties beyond my control and unrelated to whether I am making errors.



So you will have to look it up for yourself.



Almost certainly you are not serious and do not want to know how to solve the system.



Since you are aware of the term "simultaneous equation" you would have searched for that topic.



The rest of your message, after the equations, is gibberish.

Answer 5

Here we have the simultaneous solution at the point:

( (13*2-3*-3)/(4*2-3*1), (4*-3-13*1)/(4*2-3*1) ) → (35/5, -25/5) → (7,-5)



ax+by=c and dx+ey=f. Multiply the first equation through by e and the second through by b to get aex+bey=ce and bdx+bey=bf, so (ae-bd)x=ce-bf, yielding x = (ce-bf)/(ae-bd) and similarly adx+bdy=cd and adx+aey=af, yielding y=(af-cd)/(ae-bd)