A system of equations is made up of a quadratic equation (LaTeX: y_1y 1) and a linear equation (LaTeX: y_2y 2). Given the the quadratic equation: LaTeX: y_1=x^2-3x-1y 1 = x 2 − 3 x − 1 and the solutions to the system: LaTeX: \left(1,-3\right)\:and\:\left(4,3\r ight)( 1 , − 3 ) a n d ( 4 , 3 ), determine the linear equation.

starting with:    

5 + x - 12 = 2x - 7   

5 + x - 12 - x = 2x - 7 - x (subtract x from both sides)   

5 - 12 = x - 7 (combine and cancel x's from both sides)   

5 - 12 + 7 = x - 7 + 7 (add 7 to both sides)   

0 = x (add numbers together on both sides)       

so in order for the equation to be true, x must be 0.   

trying x = -0.5   

5 + x - 12 = 2x - 7 (original equation)   

5 + -0.5 - 12 = 2(-0.5) - 7 (plug in value of x)   

5 + -0.5 - 12 = -1 - 7 (did multiplication)   

-7.5 = -8 (did addition on both sides)       

since -7.5 is obviously not equal to -8, the value of -0.5 for x is wrong.   

trying x = 0   

5 + x - 12 = 2x - 7 (original equation)   

5 + 0 - 12 = 2(0) - 7 (plug in value of x)   

5 + 0 - 12 = 0 - 7 (did multiplication)   

-7 = -7 (did addition on both sides)       

since -7 equals -7, that demonstrates that the value of 0 for x is correct.   

trying x = 1   

5 + x - 12 = 2x - 7 (original equation)   

5 + 1 - 12 = 2(1) - 7 (plug in value of x)   

5 + 1 - 12 = 2 - 7 (did multiplication)   

-6 = -5 (did addition on both sides)       

since -6 is obviously not equal to -5, the value of 1 for x is wrong.

answer: the angel abc is 30°,and the abc is a triangle where the angel bac is 90°

so ac is the half of 24=12ft,and from the pythagorean theorem we find ba

ba=√bc^2-ac^2=√576-144=√432=12√3ft

so bac is 12+12√3=12(1+√3)ft

step-by-step explanation:

this can be solved first with pythagorean theorem, which tells us to square both legs of the triangle and sum it. this will be equal to the square of the hypotenuse.

the hypotenuse is the side "opposite" the 90 degree angle, which is x.

so we can write using pythagorean theorem,

simplifying:

to further simplify this radical, we need to use the property  

thus

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