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Some simultaneous questions/quadratic inequality equation question i don't understand. Pls give me the answer and the explanation. I need it for my maths homework
1) y=x-3
y=x^2-3x-8
2) y=1-x
y^2-xy=0
3) x+y=7
x^2+y^2=25
4) x^2-3x-5<0
Original post by realed
Some simultaneous questions/quadratic inequality equation question i don't understand. Pls give me the answer and the explanation. I need it for my maths homework
1) y=x-3
y=x^2-3x-8
2) y=1-x
y^2-xy=0
3) x+y=7
x^2+y^2=25
4) x^2-3x-5<0


Its worth having a read of the sticky at the top of the forum about not just asking for the answer/working. What have you tried to solve (any of) these? Using substituion, rather than elimination, would be a good thing to try for some of the questions. Upload what you tried?
(edited 1 year ago)
Original post by realed
Some simultaneous questions/quadratic inequality equation question i don't understand. Pls give me the answer and the explanation. I need it for my maths homework
1) y=x-3
y=x^2-3x-8
2) y=1-x
y^2-xy=0
3) x+y=7
x^2+y^2=25
4) x^2-3x-5<0

Like @mqb2766 said, there's little point in us giving you the answers. You won't be learning anything, and we will be doing you a disservice when it comes to the exams. The homework is there to prepare you for the exams and coursework, not to be an inconvenience.

However, I won't leave you hanging and not offer any help. If you understand the basic principles of quadratic equations, simultaneous equations, and inequalities, then you're missing the problem solving aspect of it.

Each of these problems involves at least one of the following steps:
1) move everything to one side and leave one side = 0 (a principle from quadratic equations)
2) factorise the quadratic
3) square and expand the quadratic
4) substitute one of the equation into the other (a defining feature of simultaneous equations)
5) flip the inequality

These steps won't be in order and the questions won't all involve the same order of steps. You would need to think through these questions. Maths is partly about solving problems, and you're refining your problem solving skills.
Reply 3
Original post by MindMax2000
Like @mqb2766 said, there's little point in us giving you the answers. You won't be learning anything, and we will be doing you a disservice when it comes to the exams. The homework is there to prepare you for the exams and coursework, not to be an inconvenience.

However, I won't leave you hanging and not offer any help. If you understand the basic principles of quadratic equations, simultaneous equations, and inequalities, then you're missing the problem solving aspect of it.

Each of these problems involves at least one of the following steps:
1) move everything to one side and leave one side = 0 (a principle from quadratic equations)
2) factorise the quadratic
3) square and expand the quadratic
4) substitute one of the equation into the other (a defining feature of simultaneous equations)
5) flip the inequality

These steps won't be in order and the questions won't all involve the same order of steps. You would need to think through these questions. Maths is partly about solving problems, and you're refining your problem solving skills.

thx for your advice i looked at the answers again and realised that i made a couple of mistakes with my working. With 1 you have to move everything to 1 side,Then factorise it to work out x. x-3= x^2-3x-8
0=x^2-4x-5
(x+1)(x+5)= x= -1 or -5/ y=2 or -4
For number 2 it's y=1-x
y^2-xy=0
so step 1 is to factorise the second one y(y-x)
Then you put the values in (1-x)(1-x-x)=(1-x)(1-2x)
x=1 or 0.5 so as y=1-x y is 0 or 0.5
3) With this one you have to choose a specific value.
x+y=7
x^2+y^2=25
y=7-x
(7-x)^2+x^2=2x^2-14x+49=24
2x^2-14x+24=0
x^2-7x+12=0
x^2-3x-4x+12
(x-3)(x-4)
x=3 or x=4
y=4 or y=3
However i completly don't understand what you mean when you say flip the inequality can you tell me what it means and give me an example pls
Original post by realed
thx for your advice i looked at the answers again and realised that i made a couple of mistakes with my working. With 1 you have to move everything to 1 side,Then factorise it to work out x. x-3= x^2-3x-8
0=x^2-4x-5
(x+1)(x+5)= x= -1 or -5/ y=2 or -4
For number 2 it's y=1-x
y^2-xy=0
so step 1 is to factorise the second one y(y-x)
Then you put the values in (1-x)(1-x-x)=(1-x)(1-2x)
x=1 or 0.5 so as y=1-x y is 0 or 0.5
3) With this one you have to choose a specific value.
x+y=7
x^2+y^2=25
y=7-x
(7-x)^2+x^2=2x^2-14x+49=24
2x^2-14x+24=0
x^2-7x+12=0
x^2-3x-4x+12
(x-3)(x-4)
x=3 or x=4
y=4 or y=3
However i completly don't understand what you mean when you say flip the inequality can you tell me what it means and give me an example pls

One thing Id always do for questions like these is a quick sketch of the functions.
1) is a line intersecting a quadratic
2) intersection of two lines, apart from when y=0
3) a line intersecting a circle and the solutions being well known pythagorean triplets
4) the region where a quadratic is below the x-axis. There isnt really a need to flip the inequality, but you may need to use the formula to get the roots.

Some solutions seem are ok - x values for 1 - typo? Its easy to sub back into the equations to verify. Also when listing your values, give the (x,y) pairs otherwise the 2*2 values can be combined in 4 ways, 2 of which are correct.
(edited 1 year ago)
Original post by realed
thx for your advice i looked at the answers again and realised that i made a couple of mistakes with my working. With 1 you have to move everything to 1 side,Then factorise it to work out x. x-3= x^2-3x-8
0=x^2-4x-5
(x+1)(x+5)= x= -1 or -5/ y=2 or -4
For number 2 it's y=1-x
y^2-xy=0
so step 1 is to factorise the second one y(y-x)
Then you put the values in (1-x)(1-x-x)=(1-x)(1-2x)
x=1 or 0.5 so as y=1-x y is 0 or 0.5
3) With this one you have to choose a specific value.
x+y=7
x^2+y^2=25
y=7-x
(7-x)^2+x^2=2x^2-14x+49=24
2x^2-14x+24=0
x^2-7x+12=0
x^2-3x-4x+12
(x-3)(x-4)
x=3 or x=4
y=4 or y=3
However i completly don't understand what you mean when you say flip the inequality can you tell me what it means and give me an example pls

Sorry, I wasn't thinking far ahead enough regarding the inequality, so @mqb2766 is correct. You don't need to do anything to the inequality, but you will need to know what the viable regions are (usually a graph will help) based on the inequality like he/she outlined.

He is also correct regarding the typo. Try working a little slower to avoid careless mistakes.

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