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HELP ASAP PLZZZZ
Solve the following system of equations by linear combination: 3d – e = 7 d + e = 5

There is no solution. The solution is (2, –1). There are an infinite number of solutions. The solution is (3, 2). 1 points

QUESTION 2 Solve, using linear combination. 4x + y = 5 3x + y = 3

(0, –3) (0, 5) (–3, 7) (2, –3)
QUESTION 3 Solve using linear combination. a – 2b = –2 2a + 2b = 14

Which ordered pair in the form (a, b) is the solution to the system of equations?

(4, 3) (3, 4) (1, 6) (5, 3) 1 points

QUESTION 4 Solve, using linear combination. 11x + 4y = 18 3x + 4y = 2

(2, 0) (4, –5) (2, 1) (2, –1) 1 points

QUESTION 5 Solve the following system of equations by linear combination:

2d + e = 8 d – e = 4 There is no solution. The solution is (5, –2). There are an infinite number of solutions. The solution is (4, 0).'

Respuesta :

kudzordzifrancis
QUESTION 1

The given system of equations is

[tex]3d - e = 7...eqn(1)[/tex]
[tex]d + e = 5...eqn(2)[/tex]

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

[tex]3d + d= 7 + 5[/tex]


[tex]4d = 12[/tex]


We divide through by 4 to obtain,


[tex]d = \frac{12}{4} [/tex]


[tex]d = 3[/tex]


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



[tex]3+ e = 5[/tex]


[tex]e = 5 - 3[/tex]


[tex]e = 2[/tex]


[tex] \boxed {The \: solution \: is \: (3, 2)}[/tex]



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,

[tex]4x - 3x = 5 - 3[/tex]



This implies that,

[tex]x = 2[/tex]


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

[tex]4(2) + y = 5[/tex]

[tex]8+ y = 5[/tex]


[tex]y = 5 - 8[/tex]



[tex]y = - 3[/tex]

The solution is

[tex](2,-3)[/tex]



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,

[tex]2a + a = 14 + - 2[/tex]




Simplify,

[tex]3a = 12[/tex]



Divide both sides by 3 to get,


[tex]a = 4[/tex]
Put a=4 into equation (2) to obtain,



[tex]2(4) + 2b = 14[/tex]


[tex]8 + 2b = 14[/tex]
[tex]2b = 14 - 8[/tex]


[tex]2b = 6[/tex]


[tex]b = 3[/tex]


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

[tex](4,3)[/tex]



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,


[tex]11x - 3x = 18 - 2[/tex]


[tex]8x = 16[/tex]


[tex]x = 2[/tex]


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


[tex]3(2) + 4y = 2[/tex]


This implies that,


[tex]6 + 4y = 2[/tex]


[tex]4y = 2 - 6[/tex]


[tex]4y = - 4[/tex]


[tex]y=-1[/tex]

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,

[tex]2d + d = 8 + 4[/tex]


[tex]3d = 12[/tex]



We divide both sides by 3 to get,


[tex]d = 4[/tex]


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

[tex]4 - e = 4[/tex]

[tex] - e = 4 - 4[/tex]



[tex] - e = 0[/tex]



[tex]e = 0[/tex]


The solution is

[tex](4,0)[/tex]

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