The factors of 10w2+Ilw

1. What is the solution of the system?

andreyandreev [35.5K]

Answer:

The given system of equations has solutions below:

1) The solution is (2,3)

2) The solution is ( $\frac{-8}{7}, \frac{5}{7}$ )

3) The solution is infinitely many solutions

4) No solution

Step-by-step explanation:

Given system of equation are

$-x+2y=4\hfill (1)$

$-4x+y=-5\hfill (2)$

To solve equation by using elimination method

Multiply eqn (2) into 2

$-8x+2y=-10\hfill (3)$

Now subtracting (1) and (3)

$-x+2y=4$

$-8x+2y=-10$

_________________

7x=14

x= $\frac{14}{7}$

$x=2$

Substitute x=2 in equation (1)

-2+2y=4

2y=4+2

$y=\frac{6}{2}$

y=3

Therefore the solution is (2,3)

2) Given equation is

$-2x+y=3\hfill (1)$

4y-4=x

Rewritting as below

$x-4y=-4\hfill (2)$

To solve equation by using elimination method

multiply (2) into 2

$2x-8y=-8\hfill (3)$

Adding (1) and (3)

-2x+y=3

2x-8y=-8

________

-7y=-5

$y=\frac{5}{7}$

substitute $y=\frac{5}{7}$ in (1)

$-2x+\frac{5}{7}=3$

$-2x=3-\frac{5}{7}$

$-2x=\frac{21-5}{7}$

$x=-\frac{8}{7}$

Therefore the solution is ( $\frac{-8}{7},\frac{5}{7}$ )

3) Given equation is $6x+2y=10\hfill (1)$

$3x+y=5\hfill (2)$

equation (1) can be written as

2(3x+y)=10

$3x+y=\frac{10}{2}$

3x+y=5

Therefore equations (1) and (2) are same therefore it has infinitely many solutions

4) Given equation is $-x-2y=14\hfill (1)$

$-2x-4y=12\hfill (2)$

multiply equation (1) into 2

$-2x-4y=28\hfill (3)$

To solve equation by using elimination method

subtracting (2) and (3)

-2x-4y=28

-2x-4y=12

_______

$28\neq -12$

therefore it has no solution

4 0

2 years ago

What is the slope of the line that passes through the points (-7,-4) and (-11,-2)? In simplest form

Serjik [45]

Answer:

Slope=-1/2

Step-by-step explanation:

Slope =(y2-y1) /(x2-x1)

A(-7,-4).... x1 =-7,y1 =-4

B(-11,-2)... x2=-11,y1=-2

Slope=(-2-(-4))/(-11-(-7))

Slope=(-2+4)/(-11+7)

Slope=2/(-4)

Slope=-1/2

3 0

2 years ago

2) Write the slope-intercept form of the equation of the line passing through the point (-5. -1) and perpendicular to

MAVERICK [17]

Sorry I just need points so I can have it on the phone later after work and I will have to go get it bruh was a day and a half day

8 0

2 years ago

Please help!! What Find the Error and explain why it is wrong!

Nata [24]

Answer:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

Step-by-step explanation:

First image attached

The error was done in Step E, because student did not multiply $2\cdot x -8$ by the negative sign in numerator. The real numerator in Step E should be:

$3-(2\cdot x -8)= 3-2\cdot x+8 = 11-2\cdot x$

Hence, Step E must be $\frac{2\cdot x -5}{(x+4)\cdot (x-4)}$ .

Second image attached

The error was done in Step C, because the student omitted the $2\cdot a \cdot b$ of the algebraic identity $(a+b)^{2} = a^{2}+2\cdot a\cdot b +b^{2}$ . Step C must be $5\cdot x = x^{2}+4\cdot x + 4$

And further steps are described below:

Step D

$x^{2}-x+4 = 0$