Maths-
General
Easy

Question

Use elimination to solve the system of equations.
y equals negative x squared plus 4 x plus 2
y equals 2 minus x

hintHint:

The elimination method is the process of eliminating one of the variables in the system of linear equations using addition or subtraction methods.
We are asked to solve the set of equations by the method of elimination.

The correct answer is: We are asked to solve the set of equations by the method of elimination.


    Step 1 of 3:
    The given sets of equations are:
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell y equals negative x squared plus 4 x plus 2 end cell row cell y equals 2 minus x space space space space space space space space space space space space space space end cell end table
    Eliminating the variable y from the equations, we get:
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell negative x squared plus 4 x plus 2 equals 2 minus x space space space space space end cell row cell negative x squared plus 4 x plus x plus 2 minus 2 equals 0 end cell row cell negative x squared plus 5 x equals 0 space space space space space space space space space space space space space space space space space space end cell row cell x squared minus 5 x equals 0 space space space space space space space space space space space space space space space space space space space space space end cell end table
    Step 2 of 3:
    Simplifying the equation further, we get the solutions as:
    space table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell x squared minus 5 x equals 0 space space end cell row cell x left parenthesis x minus 5 right parenthesis equals 0 space end cell row cell x equals 0 space space space space space space space space space space space space space end cell row cell x minus 5 equals 0 space space space space space space end cell row cell x equals 5 space space space space space space space space space space space space end cell end table
    Step 3 of 3:
    Substitute the values of x in any of the equation to get the solutions for y.
    When x=0,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell y equals 2 minus x end cell row cell y equals 2 minus 0 end cell row cell y equals 2 space space space space space space end cell end table
    When x=5,
    table attributes columnalign right left right left right left right left right left right left columnspacing 0em 2em 0em 2em 0em 2em 0em 2em 0em 2em 0em end attributes row cell y equals 2 minus x end cell row cell y equals 2 minus 5 end cell row cell y equals negative 3 space space space end cell end table
    Thus, the solutions are:left parenthesis 0 comma 2 right parenthesis straight & left parenthesis 5 comma negative 3 right parenthesis
    .

     
    For any equation of the form: x squared plus left parenthesis a plus b right parenthesis x plus a b, the solutions are: x equals negative a straight & x equals negative b

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