Question

Please help! thank you

Answer 1

ANSWER

1. A

2. C

3. D

4. B

5. E

EXPLANATION

QUESTION A

The given system of equation is :

$x-4y=-8...(1)$

and

$2x-3y=-16...(2)$

Make x the subject in equation (1)

This implies that;

$x=4y-8...(3)$

Put equation(3) into equation (2).

$2(4y-8)-3y=-16$

$8y-16-3y=-16$

Group like terms;

$8y-3y=-16+16$

$5y=0$

$y=0$

Put y=0 into equation(3).

$x=4(0)-8$

$x=-8$

(-8,0)

QUESTION B

First equation: $x-y=-2$

Second equation: $3x-y=0$

Subtract the first equation from the second equation.

$3x-x-y+y=0+2$

Simplify;

$2x=2$

$x=1$

Put x=1 into any of the equations, say the first one.

$1-y=-2$

$-y=-2-1$

$-y=-3$

$y=3$

(1,3)

QUESTION C

First equation: $x+y=-4$

Second equation: $-3x+2y=2$

Make y the subject in the first equation;

$y=-x-4$

Put y=-x-4 into the second equation;

$-3x+2(-x-4)=2$

Expand:

$-3x-2x-8=2$

Group like terms:

$-3x-2x=2+8$

$-5x=10$

$x=-2$

Put x=-2 into $y=-x-4$

This implies that;

$y=--2-4$

$y=-2$

(-2,-2)

QUESTION D

First equation:4x+6y=-12

Second equation: 2x+3y=9

Divide the first equation by 2;

New equation: 2x+3y=-6

Subtract the second equation from the new equation;

$2x-2x+3y-3y=-6-9$

$0=-18$. This statement is not true.

The system has no solution;

QUESTION E

First equation: $x+y=-1$

Second equation: $4x+4y=-4$

Divide the second equation by 4;

New equation: $x+y=-1$

Subtract the second equation from the new equation;

$x-x+y-y=-1+1$

$\Rightarrow 0=0$ This statement is true.

This system has infinitely many solutions.