Question

HELP ASAP PLZZZZ

Answer 1

QUESTION 1

The given system of equations is

$3d - e = 7...eqn(1)$
$d + e = 5...eqn(2)$

To solve by linear combination, we add equation (1) to equation (2) to get,

$3d + d= 7 + 5$


$4d = 12$


We divide through by 4 to obtain,


$d = \frac{12}{4}$


$d = 3$


We put d=3 into equation (2) to get,



$3+ e = 5$


$e = 5 - 3$


$e = 2$


$\boxed {The \: solution \: is \: (3, 2)}$



QUESTION 2


The given system is

4x + y = 5 ...eqn(1)

3x + y = 3 ...eqn(2)


To solve by linear combination, we subtract equation (2) from equation (1) to eliminate y from the equation.

This will give us,

$4x - 3x = 5 - 3$



This implies that,

$x = 2$


Put x=3 into equation (1) to get,

$4(2) + y = 5$

$8+ y = 5$


$y = 5 - 8$



$y = - 3$

The solution is

$(2,-3)$



QUESTION 3

We want to solve the system;


a – 2b = –2 ....eqn(1)


2a + 2b = 14...eqn(2)

by linear combination.


We need to add equation (1) to equation (2) to eliminate b.


This implies that,

$2a + a = 14 + - 2$




Simplify,

$3a = 12$



Divide both sides by 3 to get,


$a = 4$
Put a=4 into equation (2) to obtain,



$2(4) + 2b = 14$


$8 + 2b = 14$
$2b = 14 - 8$


$2b = 6$


$b = 3$


The ordered pair in the form (a, b) is

$(4,3)$



QUESTION 4

The given system of equations is


11x + 4y = 18 ...eqn(1)

3x + 4y = 2 ...eqn(2)


We subtract equation (2) from equation (1) to get,


$11x - 3x = 18 - 2$


$8x = 16$


$x = 2$


Put x=2 into equation (2) to obtain,


$3(2) + 4y = 2$


This implies that,


$6 + 4y = 2$


$4y = 2 - 6$


$4y = - 4$


$y=-1$

The correct answer is (2,-1).




QUESTION 5

The given system is ;

2d + e = 8...eqn1

d – e = 4...eqn2


We add the two equations to eliminate e.


This implies that,

$2d + d = 8 + 4$


$3d = 12$



We divide both sides by 3 to get,


$d = 4$


We put d=4 into equation (2) to get,

$4 - e = 4$

$- e = 4 - 4$



$- e = 0$



$e = 0$


The solution is

$(4,0)$