Can anyone assist me in solving these simultaneous equations?

Question

y = x + 2
y= x2

Answer 1

y = x + 2- - - - - -Equation 1
y = x²- - - - -- - - Equation 2
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Insert the y value of equation 2 into equation 1

y = x + 2

x² = x + 2

x² - x = x + 2 - x

x = 2

The answer is x = 2

Insert the x value into equation 1

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y = x + 2

y = 2 + 2

y = 4

The answer is y = 4

Insert the y value into equation 1

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Check for equation 1

y = x + 2

4 = 2 + 2

4 = 4

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Check for equation 2

y = x²

4 = (2)²

4 = 4

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The solution set is { 2, 4}

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Answer 2

y=x+2
y=2x ( both x and y should be on one side and nos on other side)

y - x = 2
y - 2x = 0 (multiply with -1) all of them
-----> -y + 2x = 0
y - x = 2 ( cancel the y's)
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x = 2 ( now u found the value of x )
y - x = 2 ( put the value of x )
y - 2 = 2
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y = 2 + 2 = 4 ( now value of y is 4)

to check : ( use anyone of all equations)
y - x = 2
4 - 2 = 2
2 = 2 ( so the values found for x & y is correct)

good luck hope u got the method


mulla sandra u r wrong

Answer 3

This reduces to X^2 = X + 2, for which the solutions are:
X=-1 or X=2. Plug in for the values for Y.

Answer 4

y=x^2 or 2x? Either way:
you can substiture in x+2 for y in y=x^2, since y=x+2, so he equation becomes x+2=x^2 or x+2=2x. For the latter equation
2+x-x=2x-x, so
2=x

For x+2=x^2
0=x^2-x-2
so if you plug in the quadratic formula
(1+/- sqrt(1+8))/2=(1+/-3)/2=2 or -1=x

Hope it helps.

Answer 5

Plug it in.

y=4 y=1
OR
x=2 x= -1

Answer 6

x^2 = x +2
x^2 -x -2 = 0
(x-2)(x+1) =0
x-2=0 or x+1 =0
x =2 then y = 4 or
x =-1 then y = 1

Answer 7

enable 3x -4y = 23 be [a] and 5x +2y = 21 be [b]. Multiply [b] by ability of two on the two factors: 2 * (5x +2y) = 2 * 21 So, 10x +4y = 40 two. enable this be [c]. upload [a] and [c] {3x -4y} + {10x +4y} = 23 + 40 two So, 13x = sixty 5 So, x = 5. placed this fee of x in [a] 3(5) -4y = 23 15 - 4y =23 -4y = 23 -15 -4y = 8 So y = -2 answer set (5, -2)