Does anyone know how do solving systems of equations by elimination ?

Eliminate a single variable to find the value of another variable.

For example,

Notice that -2x and 2x has the same "absolute" value right? -2x+2x = 0, that's how elimination work.

However, if you don't want to eliminate the "x", you can eliminate "y" instead.

Just simply multiply in.. From the equation above, we're going to eliminate y now.

However, both y in both equations are not the same (Doesn't have same coefficient) So we need to multiply in the equation that has the lesser coefficient (From both equations, the first equation has lesser coefficient)

So multiply 4 in -2x+y = 3 (To make the coefficient same.)

Now we can eliminate y but we can't eliminate x because -8x and 2x both have the different variables.

However, it's 4y and 4y, not -4y and 4y. We can only eliminate as if the absolute of both variables are the same and the value must be opposite (Absolute of both variables are 4 but the value must be opposite aka one must be negative and another must be positive.)

So one of two 4y(s) must be negative, for me I'd choose -8x+4y=12 because -8x is in negative and I want it to be positive

So multiply "minus" in -8x+4y=12

Now one is -4y, we can eliminate the y.

Then just simply eliminate y (-4y+4y = 0)

From x, 8x+2x = 10x and -12+2 = -10

10x = -10

x = -1 then substitute x in any equations but don't substitute in both equations, just one then you get the value of y.

distance covered by pamela in 9 l of fuel = 99km

let us apply unitary method here.

distance covered by pamela in 1 l of fuel =99/9 =11 km

now let us find distance covered by pamela if she has 12 l of fuel.

distance covered in 12 l = distance covered in 1 l of fuel * 12

we know that distance covered by here in 1 l is 11 km.

so distance covered in 12 l of fuel = 11*12 = 132 km.

i only know the sequence of arithmetic it adds the first is 5+4 then it goes up as it adds to then it is 9+5 then 14+6 then to 20+7 and so on. 5,9,14,20,27,35.

step-by-step explanation: