VCE Methods Question Thread!

[snowisawesome]

mx - 4y= m+3
4x + (m+10)y = -2
m is a real constant
Find the value of m for which there are infinitely many solutions
Equation 1
-4y = -mx +(m+3)
y = (mx/4) + (m+3/-4)

Equation 2
(m+10)y = -4x  - 2
y = (-4x/m+10) -(2/m+10)

For infinite solutions
m1 = m2
(-4/m+10) = (m/4)
m(m+10) = -16
m^2 + 10m +16 = 0
(m+8)(m+2) = 0
m = -8, m = -2

c1 = c2
(m+3/-4) = (-2/m+10)
(m+3)(m+10) = 8
m^2 + 13m + 30 = 8
m^2 + 13m + 22 = 0
(m+11)(m+2) = 0
m = -11, m = -2

Are these answers correct?

[Unsplash]

mx - 4y= m+3
4x + (m+10)y = -2
m is a real constant
Find the value of m for which there are infinitely many solutions
Equation 1
-4y = -mx +(m+3)
y = (mx/4) + (m+3/-4)

Equation 2
(m+10)y = -4x  - 2
y = (-4x/m+10) -(2/m+10)

For infinite solutions
m1 = m2
(-4/m+10) = (m/4)
m(m+10) = -16
m^2 + 10m +16 = 0
(m+8)(m+2) = 0
m = -8, m = -2

c1 = c2
(m+3/-4) = (-2/m+10)
(m+3)(m+10) = 8
m^2 + 13m + 30 = 8
m^2 + 13m + 22 = 0
(m+11)(m+2) = 0
m = -11, m = -2

Are these answers correct?

I'm not 100% sure, but so far you are correct. You just need to state that the value that of m for which there are infinitely many solutions is m = -2. As the lines will be the exact same when this occurs.

[snowisawesome]

I'm not 100% sure, but so far you are correct. You just need to state that the value that of m for which there are infinitely many solutions is m = -2. As the lines will be the exact same when this occurs.

What about m = -8?

[Unsplash]

What about m = -8?

If you sub in m = -8, the lines are parallel but are not the same line, and therefore have no solutions.

[snowisawesome]

If you sub in m = -8, the lines are parallel but are not the same line, and therefore have no solutions.

Thanks

[snowisawesome]

A chemical manufacturer has an order for 500 litres of a 25% acid solution. Solutions of 30% and 18% are available in stock.

The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
Let x = the amount of 30% solution required
Let y = the amount of 18% solution required
Use simultaneous equations in x and y to determine the amount of each solution required.

Equation 1
30x + 18y = 500
Not sure how to get equation 2?

[zhen]

A chemical manufacturer has an order for 500 litres of a 25% acid solution. Solutions of 30% and 18% are available in stock.

The manufacturer wishes to make up the 500 litres from a mixture of 30% and 18% solutions.
Let x = the amount of 30% solution required
Let y = the amount of 18% solution required
Use simultaneous equations in x and y to determine the amount of each solution required.

Equation 1
30x + 18y = 500
Not sure how to get equation 2?

x+y=500 (As the amounts of each solution must add to give the total volume)
0.3x+0.18y=(0.25)(500) (So you multiply out the percentage acid you have by the amount of the solution to get the amount of acid. This is equal to 25% of 500, which is 125)

[snowisawesome]

x+y=300 (As the amounts of each solution must add to give the total volume)
0.3x+0.18y=(0.25)(500) (So you multiply out the percentage acid you have by the amount of the solution to get the amount of acid. This is equal to 25% of 500, which is 125)

Where did you get the 300 from?

[zhen]

Where did you get the 300 from?

Whoops, supposed to be 500.

[snowisawesome]

It can certainly be done by looking at the distance between points, however, the first distance is not necessarily 3 (you could call it 3z for instance, and the other one would be z)

You'd also need to use the fact that the point P is on the line segment AB (i.e. find the equation of the line and substitute it in).
This, however, could be an algebraic nightmare.
A much simpler solution would be to consider the problem geometrically (there is a formula that we could generate for the coordinates of a point split in an arbitrary ratio internally on a line segment, but isn't required for methods).
The ratio AP:PB=3:1 suggests that the line segment is split into 4 equal length segments (and P is closer to B than to A).
If we let M be the midpoint of the line segment AB, then P is simply the midpoint of MB.

Could P be the midpoint of MA as well?

[jazzycab]

Could P be the midpoint of MA as well?

No. Because the line segment is split in the ratio AP:PB=3:1, P must be the midpoint of MB (P has to be closer to B than A).
Drawing a diagram may be helpful here.

[snowisawesome]

x+y=500 (As the amounts of each solution must add to give the total volume)
0.3x+0.18y=(0.25)(500) (So you multiply out the percentage acid you have by the amount of the solution to get the amount of acid. This is equal to 25% of 500, which is 125)

I don't get it - could you please explain further?

[zhen]

I don't get it - could you please explain further?

Ok, x is the amount of 30% solution. Y is the amount of 18% solution. The amount of the mixture of the two solutions is x+y. It is already confirmed that the mixture of the two solutions has a total amount 500 litres.
So x+y=500
Ok, so 0.3x is the amount of acid in the 30% solution.
0.18y is the amount of acid in the 18% solution.
You know that the amount of acid in the mixture of these two is (0.25)(500) as 25% of the solution is acid. So, 125 litres is acid. This is equal to the amount of acid in the 18% and 30% acid solutions added together.
0.3x+0.18y=125
Then solve these equations simultaneously. Is this clearer?

[snowisawesome]

Ok, x is the amount of 30% solution. Y is the amount of 18% solution. The amount of the mixture of the two solutions is x+y. It is already confirmed that the mixture of the two solutions has a total amount 500 litres.
So x+y=500
Ok, so 0.3x is the amount of acid in the 30% solution.
0.18y is the amount of acid in the 18% solution.
You know that the amount of acid in the mixture of these two is (0.25)(500) as 25% of the solution is acid. So, 125 litres is acid. This is equal to the amount of acid in the 18% and 30% acid solutions added together.
0.3x+0.18y=125
Then solve these equations simultaneously. Is this clearer?

Yes thank you!

[snowisawesome]

Does it matter which questions I do from a particular exercise?
I'm currently attempting a few chapters of chapter 2 from mathsquest and following my school's suggested questions but if the suggested questions change next year, then will it make a difference weather i do the new set of suggested questions or the old set of suggested questions?