Maths-
General
Easy

Question

y equals 2 x plus 4
y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
The system of equations above is graphed in the xy -plane. At which of the following points do the graphs of the equations intersect?

  1. ( - 3, - 2 )
  2. ( - 3, 2 )
  3. ( 5, - 2 )
  4. (5, 14 )

hintHint:

Hint:
The given two equations are graphed on the xy plane. Obviously, every point on the curve will satisfy the equation. The point at which the lines intersect will satisfy both the given equations. Such a point will be the solution of the system of equations. So, we simply need to solve the given system of equations to get the point which intersects the graphs of the equations.

The correct answer is: (5, 14 )


    The given equations are:
    y equals 2 x plus 4
    y equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We will use the method of substitution to solve these equations.
    From the first equation, we use the value of  in the second equation,
    2 x plus 4 equals left parenthesis x minus 3 right parenthesis left parenthesis x plus 2 right parenthesis
    We get an equation in one variable.
    Simplifying this, we get
    2 x plus 4 equals x left parenthesis x plus 2 right parenthesis minus 3 left parenthesis x plus 2 right parenthesis
    2 x plus 4 equals x squared plus 2 x minus 3 x minus 6
    2 x plus 4 equals x squared minus x minus 6
    Taking all the terms on one side, we have
    x squared minus x minus 2 x minus 6 minus 4 equals 0
    Thus, we get a quadratic equation,
    x squared minus 3 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    factorization to solve We use middle term the above equation.
    x squared minus 5 x plus 2 x minus 10 equals 0
    Now, taking  common from the first two terms and 2 from the last two terms, we have
    x left parenthesis x minus 5 right parenthesis plus 2 left parenthesis x minus 5 right parenthesis equals 0
    left parenthesis x minus 5 right parenthesis left parenthesis x plus 2 right parenthesis equals 0
    So, we get two solutions for  given by
    x minus 5 equals 0 not stretchy rightwards double arrow x equals 5 space of 1em text  and  end text space of 1em x plus 2 equals 0 not stretchy rightwards double arrow x equals negative 2
    Using these values of x in the first equation, we get two values of y
    text  For  end text x equals 5 text , we have  end text y equals 2.5 plus 4 equals 10 plus 4 equals 14
    text  For  end text x equals negative 2 text , we have  end text y equals 2 left parenthesis negative 2 right parenthesis plus 4 equals negative 4 plus 4 equals 0
    Thus, we get two solutions for the given system of equation
    left parenthesis 5 comma 14 right parenthesis text  and  end text left parenthesis negative 2 comma 0 right parenthesis
    We can see that ( 5, 14 )  is in the options.
    The correct option is D).

    Note:
    We can solve the equations by other methods too, but the method of substitution is most suitable here. Also, instead of solving
     x squared minus 3 x minus 10 equals 0by middle term factorisation, we could also use the quadratic formula, which is given by
    x equals fraction numerator negative b plus-or-minus square root of b squared minus 4 a c end root over denominator 2 a end fraction
    Where the standard form of equation is a x squared plus b x plus c equals 0

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