Simultaneous equations w/ linear and quadratic (substitution); e.g. x+y=8 and y=x^2

S

Simonsky

Junior Member
Joined
Jul 4, 2017
Messages
128
I though I understood the principle of substituting from the linear equation into the 'leading term' (if that's the right expression for it).

However, I'm having some basic problems which means I'm overlooking something.

here's an example:

x+y=8 and y=x^2 Should be simple! here's my working:

x+y=8 => x= 8-y so y=(8-y)^2 => 64-16y =y^2 or y^2 -16y+64 - of course I didn't need to do that step as it was already factorised as (8-y)^2

so y=8 and if x+y=8 the x =0 - this is not the answer in the book! I've gone wrong and can't see it and will go 'd'oh' (www.youtube.com/watch?v=bu-RMcvSzPw)
 
Simonsky said:
I though I understood the principle of substituting from the linear equation into the 'leading term' (if that's the right expression for it).

However, I'm having some basic problems which means I'm overlooking something.

here's an example:

x+y=8 and y=x^2 Should be simple! here's my working:

x+y=8 => x= 8-y so y=(8-y)^2 => 64-16y =y^2 or y^2 -16y+64 - of course I didn't need to do that step as it was already factorised as (8-y)^2

so y=8 and if x+y=8 the x =0 - this is not the answer in the book!
Click to expand...

I think you lost the y. The equation y=(8-y)^2 expands not to 64 - 16y = y^2, but to y = 64 - 16y + y^2, and so to y^2 - 17y + 64 = 0

What you seem to have actually solved is (8-y)^2 = 0, rather than = y, whose solution is of course y=8.

You can now solve the resulting quadratic equation in one variable, by the quadratic formula or by completing the square.

There are other ways you could have solved the system. I would have saved work by substituting for y rather than x:

y = 8 - x, so 8 - x = x^2, and x^2 + x - 8 = 0.
​

Or, I might have just plugged the second equation into the first:

x + y = 8 becomes x + x^2 = 8.
​

The result is the same.

Probably the lesson to take from this is the usual: take your time, and write out more steps that you want to, fully.
 
Dr.Peterson said:
I think you lost the y. The equation y=(8-y)^2 expands not to 64 - 16y = y^2, but to y = 64 - 16y + y^2, and so to y^2 - 17y + 64 = 0

What you seem to have actually solved is (8-y)^2 = 0, rather than = y, whose solution is of course y=8.

You can now solve the resulting quadratic equation in one variable, by the quadratic formula or by completing the square.

There are other ways you could have solved the system. I would have saved work by substituting for y rather than x:
y = 8 - x, so 8 - x = x^2, and x^2 + x - 8 = 0.
​

Or, I might have just plugged the second equation into the first:
x + y = 8 becomes x + x^2 = 8.
​

The result is the same.

Probably the lesson to take from this is the usual: take your time, and write out more steps that you want to, fully.
Click to expand...


Indeed! another 'd'oh' moment for me par excellence! I can't see what is in front of me-it's as if my brain goes rigid and a can't break away from the faulty perception! One lives in hope. Thanks.
 
You must log in or register to reply here.
Top