# Simultaneous Equations

• December 5th 2010, 07:40 AM
Ilsa
Simultaneous Equations
If (1,p) is a solution of the simultaneous equations

12x(square) - 5y(square) = 7
2p(square)x - 5y = 7

find the value of p and the other solution.
• December 5th 2010, 08:29 AM
Punch
$12x^2-5y^2=7$
$2p^2x-5y=7$

$12x^2-5y^2=2p^2x-5y$

$2x(6x-p^2)=5y[y-1]$

substitute (1,p)

$2(6-p^2)=5p[p-1]$

$12-2p^2=5p^2-5p$

$7p^2-5p-12=0$

$p=\frac{25+-\sqrt{25+336}}{14}$

$=3\frac{1}{7} or \frac{3}{7}$
• December 5th 2010, 09:42 AM
Soroban
Hello, Ilsa!

Quote:

$\text{If }(1,p)\text{ is a solution of the simultaneous equations:}$

. . . . . $\begin{array}{cccc}12x^2 - 5y^2 &=& 7 & [1] \\
2p^2x - 5y &=& 7 & [2] \end{array}$

$\text{find the value of }p\text{ and the other solution.}$

Substitute $(1,p)$ into [1]: . $12 - 5p^2 \:=\:7 \quad\Rightarrow\quad 5p^2 \:=\:5 \quad\Rightarrow\quad p \:=\:\pm1$

Substitute $(1,p)$ into [2]: . $2p^2 - 5p \:=\:7 \quad\Rightarrow\quad 2p^2 - 5p - 7 \:=\:0$

. . . . . . . . . . . . . . $(p+1)(2p-7) \:=\:0 \quad\Rightarrow\quad p \:=\:\text{-}1,\:\frac{7}{2}$

$\text{Hence: }\:\boxed{p \:=\:\text{-}1}$. . One solution is ${1,\:\text{-}1)$

The two equations become: . $\begin{array}{cccccccc}
12x^2 - 5y^2 &=& 7 & [3] \\
2x - 5y &=& 7 & \Rightarrow & y &=& \frac{2x-7}{5} & [4] \end{array}$

Substitute [4] into [3]: . $12x^2 - 5\left(\frac{2x-7}{5}\right)^2 \:=\:7$

. . which simplifies to: . $2x^2 + x - 3 \:=\:0$

. . which factors: . $(x - 1)(2x+3) \:=\:0$

. . and has roots: . $x \:=\:\text{-}1,\:\frac{3}{2}$

Substitute into [4]: . $y \:=\:\text{-}1,\:\text{-}2$

The solutions are: . $(x,y) \;=\;(1,\:\text{-}1),\;\boxed{\left(\text{-}\tfrac{3}{2},\:\text{-}2\right)}$

• December 5th 2010, 11:58 AM
Quote:

Originally Posted by Ilsa
If (1,p) is a solution of the simultaneous equations

12x(square) - 5y(square) = 7
2p(square)x - 5y = 7

find the value of p and the other solution.

$12x^2-5y^2=7$ is non-linear.

$\left(2p^2\right)x-5y=7$ is linear.

$(1,p)$ is a solution $\Rightarrow\ f(1)=p$ for both equations.

$2p^2-5p-7=0\Rightarrow\ (2p-7)(p+1)=0\Rightarrow\ p=-1,\;\;p=\frac{7}{2}$

Test these solutions for "p" in the curve equation using $f(1)=p$
as the "p" values were calculated from the line.

$12(1)-5(-1)^2=7\;\;?\;\;\;correct$

$12(1)-5\left(\frac{49}{4}\right)=7\;\;?\;\;\;incorrect$

Hence.. $p=-1$

Second solution

$2x-5y=7\Rightarrow\ 2x=5y+7,\;\;\;6x=3(5y+7)=15y+21$

$12x^2-5y^2=7=2x-5y\Rightarrow\ 12x^2-2x=5y^2-5y$

$2x(6x-1)=5y^2-5y\Rightarrow\ (5y+7)(15y+20)=5y^2-5y$

$75y^2+105y+105y+140=5y^2-5y\Rightarrow\ 70y^2+210y+140=0$

$7y^2+21y+14=0\Rightarrow\ y^2+3y+2=0\Rightarrow\ (y+1)(y+2)=0$

$y=-1,\;\;y=-2$