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Mathematics, 07.11.2019 19:31 roycallender01

Consider the equation 4xβˆ’2=βˆ’2(ax+3), with constant a. which of the following statements is true about the solution to the equation?

there is no value of a for which the equation will have no solution.

if a=2, then there is no solution.

if a=0, then there is no solution.

if a=βˆ’2, then there is no solution.

substitute βˆ’4 for a and 16 for b in the given equations. which statements are true?

select all that apply.

the equation 2(2xβˆ’11)+6=ax+b has no solution.

the equation 4xβˆ’8(xβˆ’2)=ax+b has infinitely many solutions.

the equation βˆ’5(4xβˆ’6)βˆ’14=ax+b has exactly one solution.

the equation 9x+3βˆ’13xβˆ’6=ax+b has infinitely many solutions.

the equation 10+x+2βˆ’6x=ax+b has no solution.

consider the equation 6+3(6xβˆ’7)=ax+b.

question 1
part a

for what values of a and b would the equation have infinitely many solutions?

a=

b=

part b

if a=18 and b=βˆ’13, then explain how to find the number of solutions to the equation.

use the drop-down menus to complete the statements.

substitute 18 for a and βˆ’13 for b on the right side of the equation. simplify the expression on the left side of the equation to .

the equation simplifies to . this means that the equation is true for number(s).

therefore, the equation has solution(s).

determine the number of solutions each equation has.

select no solution, one solution, or infinitely many solutions for each equation.

4(yβˆ’2)+4=2(2yβˆ’2)

no solution
no solution – 4 times open paren y minus 2 close paren plus 4 is equal to 2 times open paren 2 y minus 2 close paren

one solution
one solution – 4 times open paren y minus 2 close paren plus 4 is equal to 2 times open paren 2 y minus 2 close paren

infinitely many solutions
infinitely many solutions – 4 times open paren y minus 2 close paren plus 4 is equal to 2 times open paren 2 y minus 2 close paren
7yβˆ’2=5y+10+2y

no solution
no solution – 7 y minus 2 is equal to 5 y plus 10 plus 2 y

one solution
one solution – 7 y minus 2 is equal to 5 y plus 10 plus 2 y

infinitely many solutions
infinitely many solutions – 7 y minus 2 is equal to 5 y plus 10 plus 2 y
6y+4βˆ’4=6y+6+2

no solution
no solution – 6 y plus 4 minus 4 is equal to 6 y plus 6 plus 2

one solution
one solution – 6 y plus 4 minus 4 is equal to 6 y plus 6 plus 2

infinitely many solutions
infinitely many solutions – 6 y plus 4 minus 4 is equal to 6 y plus 6 plus 2
5(y+3)=9βˆ’y

no solution
no solution – 5 times open paren y plus 3 close paren is equal to 9 minus y

one solution
one solution – 5 times open paren y plus 3 close paren is equal to 9 minus y

infinitely many solutions
infinitely many solutions – 5 times open paren y plus 3 close paren is equal to 9 minus y

30pts for this need

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Answers: 1

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