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ok, what is a simply linear equation...?
next, what is a simple quadratic equation...?
what do I do now?
well I wouldn't have considered the 2 equations simple... the quadratic is positive definite.... which means it won't cut the x-axis by anyway, you have 2 equations... next step is to solve them simultaneously y = 5x + 3 y = 16x^2 + 4x + 2 so to solve them just equate them 5x + 3 = 16x^2 + 4x + 2 then solve for x. but you have chosen 2 difficult equations... I've attached a file, its a graph of your 2 equations
so you may want to rethink your equations.
i just made up 2 random equations I don't know what equations to think of
ok... so start with simple you linear equation is almost ok change the y-intercept to 6 so use y = 5x + 6 now a really simply quadratic is \[y = x^2\] So equating them you get \[5x + 6 = x^2\] or \[x^2 - 5x - 6 = 0\] this is a quadratic equation that can be solved by factoring... so that is what you need to do, solve for x, when you get the 2 answers, substitute the x values into either equation to find the matching y values... so you will have 2 point of intersection... if you have to graph the curves 1 use a table of values that includes both x values or just just some graphing software line https://www.desmos.com/calculator
thank you! @campbell_st