Quadratic Simultaneous Equations involving elmination?

Question

Can somebody please help me solve the following through the process of elimination.

2y - x = 3
x^2 + y^2 = 18

Another question I have is,
What is 57,000,000 cubic cm in cubic metres?

Answer 1

2y-x=3,Thus x=2y-3.Substituting this value of x in the other equation we have (2y-3)^2+y^2=18.Or 4y^2-12y+9+y^2=18.
Or 5y^2-12y-9==0.We can also write this as 5y^2-15y+3y-9=0
or 5y(y-3)+3(y-3)=0 or (5y+3)(y-3)=0.From this we get two values of y which are y=3 and the other y= -3/5.Substituting these value of y in the first eaquation,against y=3,we get x=3;
against y= -3/5,we get x= -21/5.
As regards the other one, 1 metre= 100cm.Thus we have
1cubic metre=1000000cubic centimetres.Please note that this has 6 zeros.The question is about 57000000 cubic cm which also has 6 zeros after 57.The answer is thus obviously 57 cubic metres.

Answer 2

there are a number of uncomplicated the type to resolve simultaneous equations. Substitution is the technique we are going to be employing. First with the equation y-x=3, the two make X or Y the priority; that's what ever you like. i visit make Y the priority. Y=X+3 Now, replace this into the 2nd equation x^2+y^2=29, subsequently getting x^2+ (x+3)^2=29 improve and simplify the equation, x^2+x^2+6x+9=29; subsequently getting 2x^2+6x+9=29. Make the quadratic equivalent to 0 via subtracting the 29; subsequently getting 2x^2+6x-20=0. you are able to divide via 2 to make issues uncomplicated, x^2+3x-10=0 Factorize the quadratic; (x+5)(x-2)=0 resolve for X x+5=0 and x-2=0; subsequently x is -5 or 2. those are the X coordinates, now it's time to locate the Y coordinates. employing the least difficult equation y=x+3, now replace the X coordinates into that equation to get the Y coordinates. subsequently, y (-5)+3, and y=(2)+3, do the undemanding maths and you will locate that Y is comparable to -2 and 5. subsequently the coordinates are ( -5, -2) and (2, 5)

Answer 3

Solve the first equation for x in terms of y: x = 2y - 3. Next substitute this expression for x in the second equation to eliminate the x: (2y - 3)^2 + y^ = 18. Now expand and collect like terms: 5y^2 - 12y - 9 = 0. You can use the quadratic formula to solve for y and then substitute these y values into x = 2y - 3 to find the corresponding x values.

Answer 4

From the first equation
-x=3-2y
=> x=2y-3
Putting yhe value of x in the 2nd eqn. we get,
(2y-3)^2+y^2=18
=> 4y^2-12y+9+y^2-18=0
=>5y^2-12y-9=0
=>5y^2-15y+3y-9=0
=>5y(y-3)+3(y-3)=0
=>(y-3)(5y+3)=0
Hence y=3 or -3/5
When y=3,x=6-3=3
when y= -3/5, x= -6/5-3=-4 1/5
Therefore x=3,y=3 OR
x= -4 1/5,y= -3/5
100 cm=1m
100*100*100 cu.cm= 1 cu.m
Therefore 57,000,000 cu.cm
=57,000,000/100*100*100 cu m
=57 cubic meter

Answer 5

Question 1
x = 2y - 3
(2x - 3)² + y² = 18
4x² - 12y + 9 + y² = 18
5y² - 12y - 9 = 0
(5y + 3).(y - 3) = 0
y = - 3 / 5, y = 3
Question 2
10^6 cm³ = 1 m³
57 x 10^6 cm³ = 57 m³

Answer 6

2y - x = 3
x^2 + y^2 = 18

From the first equation
x=2y-3

Substitute this value if x into the second equation.
(2y-3)^2 + y^2 = 18
4y^2 - 12y + 9 + y^2 = 18
5y^2 - 12y - 9 = 0
5y^2 - 15y + 3y - 9 = 0
5y(y-3) + 3(y-3) = 0
(y-3)(5y+3)=0

y=3 -----> x = 3
y=-3/5 ---> x= -21/5

x=3, y=3
OR
x= -21/5, y= -3/5

Check:
3^2 + 3^2 = 9+9=18
(-21/5)^2 + (-3/5)^2 = (441+9)/25 = 18

Confirmed
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1m = 100cm
1m^3 = (100cm)^3 = 1,000,000 cm^3

1,000,000 cm^3 is 1m^3
57,000,000 cm^3 is 57m^3
57 cubic metres

Answer 7

I was too late to get the quadratic part, so I'll do the conversion.

There are 100 cm in each m.

57,000,000cm^3 x (1m/100cm)^3 =

57,000,000cm^3 x 1m^3/1,000,000cm^3

= 57m^3