Simulatenous equations (Quadratic + linear)

**Drdj** · March 27th 2011, 05:56 PM

If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b

A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)

I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7

Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?

Thanks

**skeeter** · March 27th 2011, 06:10 PM

Originally Posted by Drdj

If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b

A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)

I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7

Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?

Thanks

you don't ... extraneous solutions show up when you sub them back into the original equations

for example ...

$\sqrt{x+2} = x$

$x+2 = x^2$

$0 = x^2 - x - 2$

$0 = (x - 2)(x + 1)$

$x = 2$ , $x = -1$

$x = -1$ is an extraneous solution caused by the resulting quadratic ... $\sqrt{whatever} \ge 0$

**topsquark** · March 27th 2011, 06:15 PM

Originally Posted by Drdj

If (2,1) is a solution fo the simultaneous equations,
x^2 + xy + ay = b
2ax + 3y = b

A) Find the value of a and of b. [a=1, b =7]
B) Find also the other solution [x=-7 and y = 7)

I have solved both parts, but I face a problem.
After part a), I substituted values of a and b into the equation to obtain
x^2 + xy + y = 7 and 2x + 3y = 7

Substituting x = (7-3y)/2 into the equation x^2 + xy + y = 7
I obtain the quadratic expression y^2 - 9y + 14 = 0
Hence, I get y = 7 and y =2 ; x = -7 and x = 1/2

Why do I reject y =2 and x = 1/2 combination ?
I know that substituting the above into 2x + 3y = 7 satisfies the equation
But it doesn't satisfy x^2 + xy + y = 7. How would I know?

Thanks

There are no extraneous solutions. Your quadratic is incorrect. I get $y^2 - 8y + 7 = 0$

-Dan

**Drdj** · March 27th 2011, 06:19 PM

Thanks skeeter ! So, for all quadratic equations, I have to substitute the final answers back, in order to check if both sets of equations are valid! Many thanks for your quick reply !

Originally Posted by skeeter

you don't ... extraneous solutions show up when you sub them back into the original equations

for example ...

$\sqrt{x+2} = x$

$x+2 = x^2$

$0 = x^2 - x - 2$

$0 = (x - 2)(x + 1)$

$x = 2$ , $x = -1$

$x = -1$ is an extraneous solution caused by the resulting quadratic ... $\sqrt{whatever} \ge 0$

**Soroban** · March 27th 2011, 06:49 PM

Hello, Drdj!

A simple algebra error . . .

$\text{If }(2,1)\text{ is a solution of the simultaneous equations:}$

. . $\begin{array}{c} x^2 + xy + ay \;=\;b \\ 2ax + 3y \;=\;b \end{array}$

$\text{(A) Find the values of }a\text{ and }b.\;\;[a=1,\:b =7]$

$\text{(B) Find also the other solution. }\;\;[x=-7,\:y = 7)$

$\text{I have solved both parts, but I face a problem. }$

$\text{After part (A), I substituted values of }a\text{ and }b\text{ into the equations}$
. . $\text{to obtain: }\:x^2 + xy + y \:=\: 7\:\text{ and }\:2x + 3y \:=\: 7$ . Yes!

$\text{Substituting }x \:=\:\frac{7-3y}{2}\text{ into the equation }x^2 + xy + y \:=\: 7$
. . $\text{I obtain the quadratic expression: }\:y^2 - 9y + 14 \:=\: 0$ . No

You should have had: . $\left(\dfrac{7-3y}{2}\right)^2 + \left(\dfrac{7-3y}{2}\right)\!y \,+\, y \;=\;7$

. . . . . . . . . . . . . . . $\displaystyle \frac{49-42y + 9y^2}{4} + \frac{7y - 3y^2}{2} + y \;=\;7$

Multiply by 4: . $49 - 42y + 9y^2 + 14y - 6y^2 + 4y \;=\;28$

. . which simplifies to: . $3y^2 - 24y + 21 \:=\:0$

Divide by 3: . $y^2 - 8y + 7 \:=\:0$

Factor: . $(y - 1)(y - 7) \:=\:0$

Hence, we have: . $y \;=\;1,\:7$

. . .which yields: . $x \;=\;2,\:\text{-}7$

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