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3 years ago

8

-3x+7=6x-2 What would the variable be?

1 answer:

Solnce55 [7]3 years ago

5 0

<span>-3x+7=6x-2
-3x+(-6x)=-7+(-2)
-9x=-9
x=1</span>

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X squared+ y squared = 2 y = 2x squared – 3 Which of the following describes the system?

cluponka [151]

Answer:

$x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

Step-by-step explanation:

I'm assuming the system is $\left \{ {x^2+y^2=2} \atop {y=2x^2-3}} \right.$ :

$x^2+y^2=2$

$x^2+(2x^2-3)^2=2$

$x^2+(4x^4-12x^2+9)=2$

$x^2+4x^4-12x^2+9=2$

$4x^4-11x^2+9=2$

$4x^4-11x^2+7=0$

$x^4-11x^2+28=0$

$(x^2-7)(x^2-4)=0$

$(4x^2-7)(x^2-1)=0$

$4x^2-7=0$

$4x^2=7$

$x^2=\frac{7}{4}$

$x=\pm\sqrt{\frac{7}{4}}$

$x^2-1=0$

$x^2=1$

$x=\pm1$

$y=2x^2-3$

$y=2(\pm\sqrt{\frac{7}{4}})^2-3$

$y=2({\frac{7}{4}})-3$

$y=\frac{7}{2}-3$

$y=\frac{1}{2}$

$y=2x^2-3$

$y=2(\pm1)^2-3$

$y=2(1)-3$

$y=2-3$

$y=-1$

Therefore, $x=-1,1,-\sqrt{\frac{7}{4} },\sqrt{\frac{7}{4}}$ and $y=-1,\frac{1}{2}$

The ordered pair solutions are $(-\sqrt{\frac{7}{4}},0.5)$ , $(\sqrt{\frac{7}{4}},0.5)$ , $(-1,-1)$ , and $(1,-1)$ .

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2 years ago

What is the vertex and intercepts for y=(5x-2)(2x+3)

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Answer:

vertex = $(-\frac{11}{20} ,-\frac{361}{40} )$

Intercepts for x = $(\frac{2}{5} ,0), (-\frac{2}{5} ,0)$

intercepts for y = (0,-6)

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How did your position affect the vital capacity of your lungs? Suggest an explanation for any difference you found .

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Answer:

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Step-by-step explanation:

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2 years ago

Find the two straight lines represented by x² + 5xy - 6y² = 0

Andreyy89

By factorizing, we have

$x^2 + 5xy - 6y^2 = (x - y) (x + 6y) = 0$

so that

$x - y = 0 \implies \boxed{y = x}$

or

$x + 6y = 0 \implies \boxed{y = -\dfrac x6}$

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I need help please :’( Simplify -4x - x

timofeeve [1]

Answer:

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Step-by-step explanation:

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