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Systems of Nonlinear Equations

Systems of Nonlinear Equations

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2

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Question 1 of 2

Solve for

y

$y$ .

y = 4 x^{2} - 6 x

$y=4x^2-6x$

y = 2 x

$y=2x$

$x_{1} =$ $x_1 =$ (0, 2)

$x_{2} =$ $x_2 =$ (2, 0)

$y_{1} =$ $y_1 =$ (0, 4)

$y_{2} =$ $y_2 =$ (4, 0)

Question 2 of 2

2. Question
Solve for $y$ .

$y=6x-3$

$y=x^2-8x+46$
- $x =$ (7)
  
  $y =$ (39)
Hint

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First, label the two equations $1$ and $2$ respectively.

$y$ $=$ $6x-3$ Equation $1$

$y$ $=$ $x^2-8x+46$ Equation $2$

Set Equations $1$ and $2$ equal to each other.

$6x-3$ $=$ $x^2-8x+46$

$6x-3$ $-6x+3$ $=$ $x^2-8x+46$ $-6x+3$ Add $-6x+3$ to both sides

$0$ $=$ $x^2-14x+49$ Simplify

$x^2-14x+49$ $=$ $0$

Since the equation is in standard form $($ $a$ $x^2+$ $b$ $x+$ $c$ $=0)$ we can factorise using the cross method.

$x^2$ $-14$ $x+$ $49$ $=0$

To factorise, we need to find two numbers that add to $-14$ and multiply to $49$

Two $-7$ ’s fit both conditions

$-7 + (-7)$ $=$ $-14$

$-7 xx -7$ $=$ $49$

[image of cross method]

Read across to get the factors.

$(x-7)(x-7)=0$

Solve for $x$ .

$x$ $-7$ $=$ $0$

$x$ $-7$ $+7$ $=$ $0$ $+7$

$x$ $=$ $7$

Now, substitute the value of $x$ to solve for $y$

$y$ $=$ $6$ $x$ $-3$

$y$ $=$ $6$ $(7)$ $-3$ $x=7$

$y$ $=$ $42-3$

$y$ $=$ $39$

$x=7; y =39$

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