Solve the system of equations two ways: 3x+y=5 and x-2y=4 A) using substitution B) using elimination

1 answer:

5 0

A. solve for 1 variable
let's solve for x in 2nd equation
add 2y to both sides
x=2y+4
sub 2y+4 for x in other equation
3(2y+4)+y=5
6y+12+y=5
7y+12=5
minu12 both sides
7y=-7
divide 7
y=-1
sub back
x=2y+4
x=2(-1)+4
x=-2+4
x=2
(2,-1)

B. eliminate
eliminate y's
multiply first equation by 2 and add to first

6x+2y=10
x-2y=4 +
7x+0y=14

7x=14
divide by 7
x=2
sub back
x-2y=4
2-2y=4
minus 2
-2y=2
divide -2
y=-1
(2,-1)

(2,-1) is answer

You might be interested in

How do I find: If 6^x=1/216,find the value of x with and without a Casio calculator?

san4es73 [151]

Answer:

The value of x is -3.

Step-by-step explanation:

To find x, we have to put everything in base 6 and continue to simplify. Since we know that 216 is 6^3, we can replace that.

6^x = 1/216

6^x = 1/(6^3)

Now, when there is a power in the denominator, we know we can move it to the numerator by making it negative. We can do that to get the bases to match.

6^x = 1/(6^3)

6^x = 6^-3

Now that they are both put in simple base 6 systems, we can just eliminate the bases and assume that the powers are equal. This lets us know that x = -3.

8 0

2 years ago

Find the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Eva8 [605]

The coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4) is $(\frac{-17}{5} ,\frac{2}{5} )$

Explanation :

the coordinates of point Q that is 2/3 of the way along the directed segment from R(-7,-2) to S(2,4)

Apply section formula to find coordinates of point Q

$(\frac{mx_2+nx_1}{m+n} , \frac{my_2+ny_1}{m+n} )$

Ratio m:n is 2:3 and point R is (x1,y1) , point S is (x2,y2)

Substitute all the values inside the formula

$(\frac{2(2)+3(-7)}{2+3} , \frac{2(4)+3(-2)}{2+3} )\\\\(\frac{-17}{5} ,\frac{2}{5} )$

The coordinates of point Q is $(\frac{-17}{5} ,\frac{2}{5} )$

Learn more : brainly.com/question/13878373

7 0

2 years ago

Find the coordinates of the midpoint of a line segment with the given endpoints (14,-8), (12,-1)

Flauer [41]

Answer:

The coordinates of the midpoint of a line segment with the given endpoints (14,-8), (12,-1) are $\mathbf{(13,\frac{-9}{2})}$

Step-by-step explanation:

We need to find the coordinates of the midpoint of a line segment with the given endpoints (14,-8), (12,-1)

The midpoint of line segment can be found using formula:

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

We have $x_1=14, y_1=-8, x_2=12, y_2=-1$

Putting values and finding midpoint

$Midpoint=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})\\Midpoint=(\frac{14+12}{2},\frac{-8-1}{2})\\Midpoint=(\frac{26}{2},\frac{-9}{2})\\Midpoint=(13,\frac{-9}{2})$