what is the result of substituting for y in the bottom equation y=x+3 y=x^2+2x-4

Question

what is the result of substituting for y in the bottom equation
y=x+3
y=x^2+2x-4

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Mary 3 weeks 2021-09-27T03:45:36+00:00 1 Answer 0

Answers ( )

    0
    2021-09-27T03:47:03+00:00

    The solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

    Solution:

    Given that,

    y = x + 3 ------- eqn 1\\\\y = x^2 + 2x - 4  ----- eqn 2

    We have to substitute eqn 1 in eqn 2

    x + 3 = x^2 + 2x - 4

    \mathrm{Switch\:sides}\\\\x^2+2x-4=x+3\\\\\mathrm{Subtract\:}3\mathrm{\:from\:both\:sides}\\\\x^2+2x-4-3=x+3-3\\\\\mathrm{Simplify}\\\\x^2+2x-7=x\\\\\mathrm{Subtract\:}x\mathrm{\:from\:both\:sides}\\\\x^2+2x-7-x=x-x\\\\\mathrm{Simplify}\\\\x^2+x-7=0

    \mathrm{Solve\:with\:the\:quadratic\:formula}\\\\\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\\mathrm{For\:}\quad a=1,\:b=1,\:c=-7\\\\x =\frac{-1\pm \sqrt{1^2-4\cdot \:1\left(-7\right)}}{2\cdot \:1}

    x = \frac{-1 \pm \sqrt{ 1 + 28}}{2}\\\\x = \frac{ -1 \pm \sqrt{29}}{2}

    x = \frac{ -1 \pm 5.385 }{2}\\\\We\ have\ two\ solutions\\\\x = \frac{ -1 + 5.385 }{2}\\\\x = 2.1925

    Also\\\\x = \frac{ -1 - 5.385 }{2}\\\\x = -3.1925

    Substitute x = 2.1925 in eqn 1

    y = 2.1925 + 3

    y = 5.1925

    Substitute x = -3.1925 in eqn 1

    y = -3.1925 + 3

    y = -0.1925

    Thus the solutions are (2.1925, 5.1925) and (-3.1925, -0.1925)

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45:7+7-4:2-5:5*4+35:2 =? ( )