Solve the system of linear equations by elimination. 4x+3y= -5 -x+3y= -10

Mathematics

1 answer:

Georgia [21]3 years ago

5 0

$4x + 3y = - 5 \\ x - 3y = 10 \\ \\ 5x = 5 \\ x = 1 \\ \\ 4 + 3y = - 5 \\ \: \: 3y = - 9 \\ y = - 3 \\ \\ (1 \: - 3)$

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galben [10]

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6 0

3 years ago

Read 2 more answers

Bonjour pourriez vous m'aider avec cette équation du second degré? j'ai du mal à comprendre le concept. merci

olga_2 [115]

If a quadratic equation has solutions, the standard procedure will always work: given an equation like

$ax^2+bx+c=0$

The solutions (if any) are given by

$x_{1,2} = \dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

In your case, you have

$a=-9,\ b=0,\ c=1$

So the formula becomes

$x_{1,2} = \dfrac{\pm\sqrt{36}}{18}= \pm\dfrac{6}{18}=\pm\dfrac{1}{3}$

Anyway, this is quite a special case, because you're missing the linear term (since $b=0$ )

This means that you can solve equations like these more easily: rearrange the equation to make it look like $ax^2=c$ . In your case, it becomes

$9x^2=1$

Divide both sides by 9:

$x^2=\dfrac{1}{9}$

Now, you know that the square of a number equals 1/9. By definition, it means that this number is the square root of 1/9. Nevertheless, both the square root and its opposite are solutions of the equation, because the minus sign will cancel out when squaring.

So, in general, you have

$ax^2=c \iff x^2 = \dfrac{c}{a} \iff x=\pm\sqrt{\dfrac{c}{a}$