1. Simultaneous Equations

I don't usually have any problem with simultaneous equations but whilst revising for a test, I came across two types that I haven't seen before and the book I have only gives answers but no explanations of how it got those answers. I've tried in various ways... but my answers never seem to match up.

Could anyone solve these two simultaneous equations step-by-step so I could see how it's done?

Q1)
$y = 1 - 4x$
$y = -4x^2$

Q2)
$xy = 9$
$x - 2y = 3$

If anyone could I would really appreciate it, thanks. And Hi to everyone on the forum - just found this forum and it looks great.

2. Hello, PaulSelb!

Welcome aboard!

These are not linear equations.
. . The recommended method is Substitution.

$1)\;\;\begin{array}{cccc}y &=& 1 - 4x &{\color{blue}[1]} \\
y &= &-4x^2& {\color{blue}[2]}\end{array}$

Equation [1] is already solved for $y\!:\;\;y \:=\:1-4x$

Substitute into [2]: . $1 - 4x \:=\:-4x^2 \quad\Rightarrow\quad 4x^2 - 4x + 1 \:=\:0$

. . $(2x-1)^2 \:=\:0 \quad\Rightarrow\quad 2x-1 \:=\:0 \quad\Rightarrow\quad x \:=\:\frac{1}{2}$

Substitute into [1]: . $y \:=\:1 - 4\left(\frac{1}{2}\right) \:=\:-1$

Solution: . $x \:=\:\frac{1}{2},\;y \:=\:-1$

$2);\;\begin{array}{cccc}xy &=& 9 & {\color{blue}[1]} \\
x - 2y &=& 3 & {\color{blue}[2]}\end{array}$

Solve [2] for $x\!:\;\;x \:=\:2y+3\;\;{\color{blue}[3]}$

Substitute into [1]: . $(2y+3)y \:=\:9 \quad\Rightarrow\quad 2y^2 + 3y - 9 \:=\:0$

Factor: . $(2y - 3)(y + 3)\:=\:0\quad\Rightarrow\quad y \:=\:\frac{3}{2},\;-3$

Substitute into [3]: . $\begin{Bmatrix}x &=& 2\left(\frac{3}{2}\right) + 3 & = & 6 \\ \\[-3mm] x &=&2(-3) + 3 &=&-3 \end{Bmatrix}$

Solutions: . $(x,y) \;=\;\left(6,\:\frac{3}{2}\right),\;(-3,\:-3)$

3. Thanks, great explanation . Got my test in a few hours so hopefully if one of these questions turns up I shouldn't have any problems.

4. I had a slightly different type in the exam the other day (possibly solved in the same way)... but I got it wrong

If anyone reads this thread again could you let me know how to solve this one?

$k(k - m) = 12$
$k(k + m) = 60$

Sorry for the easy stupid questions, I really enjoy math and want to understand bits more!

EDIT: I think I did it may have done it actually...

$k^2 - km = 12$
$k^2 + km = 60$
$k+m = \frac{60}{k}$
$m = \frac{60}{k} - k$

$k^2 - k(\frac{60}{k} - k) = 12$

$2k^2 - 60 = 12$

$2k^2 - 72 = 0$

$(2k - 12)(k + 6) = 0$

$k = \pm 6$

$36 + \pm6(m) = 60$

$m = \frac{24}{\pm6}$

$m = \pm 4$

I think that's about right anyway...