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try using y = 2x + 1 y = x^2 - x - 6
General form of a linear equation is: ax + by + c = 0 General form of a Quadratic equation is: a(x^2) + bx + c = 0
ok basically like @welshfella said
how would i solve algebraically though?
Equate both equtions: 2x + 1 = x^2 - x - 6 Now arrange it...by combining like terms...
I'm not sure/// comine 2x-x?
how can you combine you put the stuff on the other side of the equal sign
x^2 - x- 2x - 6-1 = 0 x^2 - 3x - 7 = 0
Now you can solve this quadratic equation using QUadratic Formula...
wait can you tell me what we have doen so far and wha i should put ?
Solve this system of equations algebraically: y = x2 - x - 6 (quadratic equation in one variable of form y = ax2 + bx + c ) y = 2x - 2 (linear equation of form y = mx + b)
Substitute from the linear equation into the quadratic equation and solve. y = x2 - x - 6 2x - 2 = x2 - x - 6 2x = x2 - x - 4 0 = x2 - 3x - 4 0 =(x - 4)(x + 1) x - 4 = 0 OR x + 1 =0 x = 4 x = -1
Find the y-values by substituting each value of x into the linear equation. y = 2(4) - 2 = 6 POINT (4,6) y = 2(-1) - 2 = -4 POINT (-1,-4)
oh so that's part 1?
that makes sense srry i was a little confused
what would i do after>
Now you can plot the points graphically and check...
ok :) ty!
wait @midhun.madhu1987 what do they mean how the solution graphically ?
https://www.desmos.com/screenshot/rljjhtzr7b to graph you plot points for a range of x and y the above link shows the graphs where they intersect are the solutions
oh so thats what i should use ?
as you see they intersect at (4,6) and (-1,-4) which agrees with the values you got earlier. I'm not sure Desmos Graphing is a very good piece of software but i dont know if your teacher will allow it.
He/she might only accept a drawing on graph paper.
yes he will i always use it :)
yep could you help with another one ?