How do you solve the system y= 1/2x + 1y=12x+1 and y= -2x + 6y=−2x+6?

2 Answers
Ratnesh Bhosale
Mar 3, 2017

x=3x=3,y=0y=0
Sol^n= (3,0)Soln=(3,0)

Explanation:

y= 1/2x+1y=12x+1

2y=x+22y=x+2----------(i)

y=-2x+6y=−2x+6---------(ii)

Substitute eq ii in eq i

2(-2x+6)=-2x+62(−2x+6)=−2x+6

-4x+12=-2x+6−4x+12=−2x+6

2x=62x=6

ul(x=3)

Put x=3 in eq ii

y=-6+6

ul(y=0)

Sol^n= (3,0)

Jim G.
Mar 3, 2017

2,2)

Explanation:

color(red)(y)=1/2x+1to(1)

color(red)(y)=-2x+6to(2)

Since both equations are expressed with color(red)(y) as the subject, we can equate the right sides.

rArr1/2x+1=-2x+6

To eliminate the fraction, multiply ALL terms on both sides by 2, the denominator of the fraction.

(cancel(2)^1xx x/cancel(2)^1)+(2xx1)=(2xx-2x)+(2xx6)

rArrx+2=-4x+12

add 4x to both sides.

x+4x+2=cancel(-4x)cancel(+4x)+12

rArr5x+2=12

subtract 2 from both sides.

5xcancel(+2)cancel(-2)=12-2

rArr5x=10

divide both sides by 5

(cancel(5) x)/cancel(5)=10/5

rArrx=2

To find the corresponding value of y, substitute x = 2 into either (1) or (2). I've chosen equation (2)

x=2toy=(-2xx2)+6=-4+6=2

"solution is "(2,2)

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