Solve for x and y. y=6x+4 y=4x-2

2 answers:

4 0

$\left\{\begin{array}{ccc}y=6x+4&(1^o)\\y=4x-2&(2^o)\end{array}\right\\\\\\subtitute\ (1^o)\ to\ (2^o):\\\\6x+4=4x-2\ \ \ \ |subtract\ 4\ from\ both\ sides\\6x=4x-6\ \ \ \ |subtract\ 4x\ from\ both\ sides\\2x=-6\ \ \ \ |divide\ both\ sides\ by\ 2\\\boxed{x=-3}\\\\subtitute\ value\ of\ "x"\ to\ the\ (1^o):\\y=6(-3)+4=-18+4\to \boxed{y=-14}$

$Answer:\boxed{ \left \{ {{x=-3} \atop {y=-14}} \right}$

ad-work [718]2 years ago

3 0

Since both are already solved for y, we can take each equation and set them equal to each other.

6x+4=4x-2
6x-4x=2x (take 4x away from both sides)
-2-4=-6 (take 4 away from both sides)
2x=-6
-6/2=-3 (divide both sides by 2)
x=-3

we then plug -3 in for an x in one of the equations to get what y is equal to.
y=6(-3)+4
y=-18+4
y=-14

So, your answers are: y=-14 and x=-3

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Match the circle equations in general form with their corresponding equations in standard form. Not all will be used.

Xelga [282]

The standard form of the equation of a circumference is given by the following expression:

 $(x-h)^{2}+(y-k)^{2}=r^{2} \\ \\ where \ (h, k) \ is \ the \ center \ of \ the \ circumference \ and \ r \ the \ radius$

On the other hand, the general form is given as follows:

 $x^{2}+y^{2}+Dx+Ey+F=0 \\ \\ where: \\ D=-2h, \ E=-2k, \ F=h^{2}+k^{2}-r^{2}$ 

In this way, we can order the mentioned equations as follows:

Equations in Standard Form:

 $\bold{a)} \ (x-6)^{2}+(y-4)^{2}=56 \\ \bold{b)} \ (x-2)^{2} + (y+6)^{2}=60 \\ \bold{c)} \ (x+2)^{2}+(y+3)^{2}=18 \\ \bold{d)} \ (x+1)^{2}+(y-6)^{2}=46$

Equations in General Form:

$\bold{1)} \ x^{2}+y^{2}-4x+12y-20=0 \\ \bold{2)} \ x^{2}+y^{2}+6x-8y-10=0 \\ \bold{3)} \ 3x^{2}+3y^{2}+12x+18y-15=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 3, \ the \ equation \ becomes: \\ x^{2}+y^{2}+4x+6y-5=0 \\ \\ \bold{4)} \ 5x^{2}+5y^{2}-10x+20y-30=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 5, \ the \ equation \ becomes: \\ x^{2}+y^{2}-2x+4y-6=0 \\ \\ \bold{5)} \ 2x^{2}+2y^{2}-24x-16y-8=0 \\ \\ If \ we \ divide \ this \ equation \ by \ 2, \ the \ equation \ becomes: \\ x^{2}+y^{2}-12x-8y-4=0$

$\bold{6)} \ x^{2}+y^{2}+2x-12y$

So let's match each equation:

$\bold{From \ a)} \\ \\ (h,k)=(6,4),\ r=2\sqrt{14} \\ D=-12, \ E=-8 \\ F=-4$

Then, its general form is:

$x^{2}+y^{2}-12x-8y-4=0$

First. a) matches 5) 

$\bold{From \ b)} \\ \\ (h,k)=(2,-6),\ r=2\sqrt{15} \\ D=-4, \ E=12 \\ F=-20$

Then, its general form is:

$x^{2}+y^{2}-4x+12y-20=0$

Second. b) matches 1) 

$\bold{From \ c)} \\ \\ (h,k)=(-2,-3),\ r=3\sqrt{2} \\ D=4, \ E=6 \\ F=-5$

Then, its general form is:

$x^{2}+y^{2}+4x+6y-5=0$

Third. c) matches 3)

$\bold{From \ d)} \\ \\ (h,k)=(-1,6),\ r=\sqrt{46} \\ D=2, \ E=-12, \ F=-9$

Then, its general form is: $x^{2}+y^{2}+2x-12y-9=0$

Fourth. d) matches 6)

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