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Andru [333]
3 years ago
5

Solve the following system of equations algebraically: y = x^2+7x−4 y = −x−4

Mathematics
2 answers:
oksian1 [2.3K]3 years ago
7 0
Y = x² + 7x - 4
y = -x - 4

x² + 7x - 4 = -x - 4
      + x       + x
x² + 8x - 4 = -4
           + 4  + 4
     x² + 8x = 0
x(x) + x(8) = 0
    x(x + 8) = 0
x = 0  or  x + 8 = 0
                   - 8  - 8
                     x = -8

      y = -x - 4
      y = -0 - 4
      y = 0 - 4
      y = -4
(x, y) = (0, -4)
         or
       y = -x - 4
       y = -(-8) - 4
       y = 8 - 4
       y = 4
 (x, y) = (-8, 4)

The solutions are (0, -4) and (8, -4).
rodikova [14]3 years ago
5 0
Y=x²+7x-4
y=-x-4

We can solve this system of equation by equalization method.
x²+7x-4=-x-4
x²+7x+x-4+4=0
x²+8x=0
x(x+8)=0
we have two linear equations :
x=0                                ⇒  x=0        ⇒  y=-x-4=-0-4=-4
(x+8)=0                         ⇒  x=-8      ⇒  y=-x-4=-(-8)-4=8-4=4
 
We have two solutions:
solution₁:  x=0;  y=-4
solution₂:  x=-8; y=4
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. Use the quadratic formula to solve each quadratic real equation. Round
Liono4ka [1.6K]

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A. No real solution

B. 5 and -1.5

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Step-by-step explanation:

The quadratic formula is:

\begin{array}{*{20}c} {\frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} \end{array}, with a being the x² term, b being the x term, and c being the constant.

Let's solve for a.

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {5^2 - 4\cdot1\cdot11} }}{{2\cdot1}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {25 - 44} }}{{2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 5 \pm \sqrt {-19} }}{{2}}} \end{array}

We can't take the square root of a negative number, so A has no real solution.

Let's do B now.

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {7^2 - 4\cdot-2\cdot15} }}{{2\cdot-2}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {49 + 120} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm \sqrt {169} }}{{-4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 7 \pm 13 }}{{-4}}} \end{array}

\frac{7+13}{4} = 5\\\frac{7-13}{4}=-1.5

So B has two solutions of 5 and -1.5.

Now to C!

\begin{array}{*{20}c} {\frac{{ -(-44) \pm \sqrt {-44^2 - 4\cdot4\cdot121} }}{{2\cdot4}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm \sqrt {1936 - 1936} }}{{8}}} \end{array}

\begin{array}{*{20}c} {\frac{{ 44 \pm 0}}{{8}}} \end{array}

\frac{44}{8} = 5.5

So c has one solution: 5.5

Hope this helped (and I'm sorry I'm late!)

4 0
2 years ago
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