Simultaneous Equations

Question 1

How to solve simultaneous equations? (Substitution with graphs)

Answer 1

1) substitute values
2) draw it in a graph or a table
3) do the other equation
4) also draw that into a graph or table
5) write the coordinates at which they intersect or if the question asks say what x is and then what y is

Question 2

How to solve simultaneous equations? (Elimination)

Answer 2

1) firstly figure out if u need to add or subtract
2) by doing this u should be just left with something x or y= something
3) make it so it’s just x or y by dividing (or multiply,adding or subtracting depending on the question)
4) substitute what u get into the equation to either get y or x

Question 3

How do u know when to add or subtract while using the elimination method?

Answer 3

If it’s the same numbers but positive and negative (eg. -3y and 3y) then add
If it’s different numbers but both are positive or negative (eg. 5x and 2x) then subtract

Question 4

How do u know whether to use substitution (not the actual method) or elimination?

Answer 4

Elimination if something is equal to a number
Substitution if it’s y= or x=

Question 5

How do use the elimination method if the pairs of numbers aren’t the same?

Answer 5

1) if u can just increase one of the equations easily then increase all of the numbers by the same amount (eg. x+y=18 and 3x+3y=6. The x+y=18 —> 3x+3y=54

2) however if that doesn’t work, find a common multiple and multiply both by their respective amounts (eg. 2x+4y=16 and 3x-5y=2. The 2x+4y=16 —>3—> 6x+12y=48. The 3x-5y=2 —>2—> 6x-10y=4

3)finally u can decide whether or not u want to add or subtract (doesn’t matter)

4) substitute to find the other thing

Question 6

How does substitution work?
Use 2x+3y=40 and y=2x

Answer 6

1) substitute the bit where it mentions x or y (depending on the 2nd equation) in the first equation —> 2x+3(2x)=40 (REMEMBER TO PUT IT IN BRACKETS as it is not always just 1 number)
2) expand the bracket —> 2x+6x=40
3) simplify —> 8x=40
4) work out the answer —> x= 5
5) substitute the answer u got to find the other x/y in the 2nd equation—> y=2*4
6) work out the answer —> y = 8

Question 7

When do u use substitution?

Answer 7

When u get the equation and the 2nd one says what y/x is in terms of x/y
Eg. 5x+2y=44 and y=x+8

The answer is x=4 and y=12 btw

Question 8

How to solve a linear and quadratic simultaneous equation? (…= number)
Eg.x^2+y^2=100 and x+y=14

Answer 8

1) make sure it’s y=… , if it’s not then rearrange it so it is —> x+y=14 —> -x —> y=14-x
2) substitute the answer for y into the equation, if it’s y^2 then write them as 2 separate brackets—> x^2+(14-x)(14-x)=100
3) expand the 2 brackets using the grid method —> 1414=196, 14-x=-14x, -x14=-14x, -x-x= x^2 (negative and negative cancel each other out)
4) replace the brackets by the expanded version —> x^2 -14x -14x +x^2 =100
5) simplify it —> 2x^2 -28x +196 =100
6) make it so that the equation is equal to 0 —> 2x^2 -28x +196 =100 —>-100 —> 2x^2 -28x +96 = 0
7) see if they all have a common factor to divide by to simplify —> 2x^2 -28x +96 = 0 —> /2 —> x^2 -14x +48 = 0
8) solve it like u would a normal quadratic equation, so find 2 numbers that adds to make the number before x and multiply to make the normal number —> -6 + -8=-14, -6*-8=48
9) write that into brackets —> (x-6)(x-8)=0
10) find what x is (flip whether it’s positive or negative)—> x= 6 or 8
11) to find what y is substitute both the numbers into the 2nd equation—> 6+y=14, 8+y=14.so when x is 6 y is 8, when x is 8, y is 6

Question 9

How to solve a linear and quadratic simultaneous equation? (y = …)
Eg. Y=x^2+8x+19 and y=2x+14

Answer 9

1) substitute the y with what u know y is equal to —> 2x+14=x^2+8x+19
2) rearrange the equation so that it is equal to 0 —> 2x+14=x^2+8x+19 —>-2x —> 14=x^2+6x+19 —> -14 —> 0= x^2+6x+5
3) turn the equation into the brackets version by finding what 2 numbers add to make the number before x and also multiply to make the normal number —> 1+5=6 and 1*5=5 so (x+1)(x+5) = 0
4) find what x is (flip whether it’s positive or negative)—> x= -1 or -5
5) to find what y is substitute both the numbers into the 2nd equation—> y= 2(-1)+14 and y=2(-5)+14. so when x is -1 y is 12, when x is -5, y is 4