systems of equations and inequalities solving systems by elimination {y= -x+1 {y= 4x-14

Question

systems of equations and inequalities
solving systems by elimination
{y= -x+1
{y= 4x-14

in progress 0
2 weeks 2021-09-08T13:32:37+00:00 2 Answers 0

Answers ( )

  1. Emma
    0
    2021-09-08T13:34:09+00:00

    Answer:

    x = 3  , y = -2

    Step-by-step explanation:

    Solve the following system:

    {y = 1 – x | (equation 1)

    y = 4 x – 14 | (equation 2)

    Express the system in standard form:

    {x + y = 1 | (equation 1)

    -(4 x) + y = -14 | (equation 2)

    Swap equation 1 with equation 2:

    {-(4 x) + y = -14 | (equation 1)

    x + y = 1 | (equation 2)

    Add 1/4 × (equation 1) to equation 2:

    {-(4 x) + y = -14 | (equation 1)

    0 x+(5 y)/4 = (-5)/2 | (equation 2)

    Multiply equation 2 by 4/5:

    {-(4 x) + y = -14 | (equation 1)

    0 x+y = -2 | (equation 2)

    Subtract equation 2 from equation 1:

    {-(4 x)+0 y = -12 | (equation 1)

    0 x+y = -2 | (equation 2)

    Divide equation 1 by -4:

    {x+0 y = 3 | (equation 1)

    0 x+y = -2 | (equation 2)

    Collect results:

    Answer: {x = 3  , y = -2

  2. Emma
    0
    2021-09-08T13:34:15+00:00

    Answer:

    x = 3, y = -2 ⇒ (3, -2)

    Step-by-step explanation:

    \left\{\begin{array}{ccc}y=-x+1\\y=4x-14&\text{chnge the signs}\end{array}\right\\\\\underline{+\left\{\begin{array}{ccc}y=-x+1\\-y=-4x+14\end{array}\right}\qquad\text{add both sides of the equations}\\.\qquad0=-5x+15\qquad\text{add}\ 5x\ \text{to both sides}\\.\qquad5x=15\qquad\text{divide both sides by 5}\\.\qquad\dfrac{5x}{5}=\dfrac{15}{5}\\.\qquad\boxed{x=3}\\\\\text{Substitute the value of}\ x\ \text{to the first equation:}\\\\y=-3+1\\\boxed{y=-2}

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