Im finding the coordinates of the point of intersection of the lines y=3x+1 and x+3y=6.
so far i substituted 1 into 2,
x+3(3x+1)=6
x+9x+3=6
10x^2=3
x=3/10 _ but as i have to then put the x value back into the equation to find y i dont think i have folowed this through correctly , can anyone put me right?

Also i have done;
x^2+2x<3
x+x<1.5
2x<1.5
x<1.5/2 < could this work, or have i also miscalculated? lol

Thnk x

2. You have found the correct value for x, so just plug that value into one of the equations to find your y coordinate. The answer should be (0.3 , 1.9)

As for the second equation, you lost your square somewhere... check your work again...you lost it in your second step...

3. Originally Posted by Lyrixa
As for the second equation, you lost your square somewhere... check your work again...you lost it in your second step...
In case you missed where you went wrong...

$\displaystyle x^2+2x<3$
$\displaystyle x^2+2x-3<0$
$\displaystyle (x+3)(x-1)<0$

Critical points, x = -3, x = 1
positive values above 1 give way to values > 0, x values between -3 and 1 are less than 0, and negative values less than -3 make way to values > 0, so x must lie between -3 and 1.