I can't see a list but any equation which is a multiple of the given equation would do.
For example we could multiply each term in the equation by 2 to give:-
226x = 106y - 4.
Answer:
B.[tex]22x=10y-12[/tex]
D.[tex]11x=5y-6[/tex]
Step-by-step explanation:
We are given that an equation
[tex]\frac{11}{3}x=\frac{5}{3}y-2[/tex]
We have to find the equation that combines with the given equation to create a system of linear equations with infinitely many solutions.
Condition of infinitely many solutions:
[tex]a_1x+b_1y+c_1=0[/tex]
[tex]a_2x+b_2y+c_2=0[/tex]
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}[/tex]
A.[tex]33x=15y-16[/tex]
[tex]\frac{11}{3}x-\frac{5}{3}y+2=0[/tex]
[tex]33x-15y+16=0[/tex]
[tex]a_1=\frac{11}{3},b_1=-\frac{5}{3},c_1=2[/tex]
[tex]a_2=33,b_2=-15,c_2=16[/tex]
[tex]\frac{a_1}{a_2}=\frac{\frac{11}{3}}{33}=\frac{1}{9}[/tex]
[tex]\frac{b_1}{b_2}=\frac{-\frac{5}{3}}{-15}=\frac{1}{9}[/tex]
[tex]\frac{c_1}{c_2}=\frac{2}{16}=\frac{1}{8}[/tex]
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq \frac{c_1}{c_2}[/tex]
Hence, it is false.
B.[tex]22x=10y-12[/tex]
[tex]22x-10y+12=0[/tex]
[tex]\frac{a_1}{a_2}=\frac{\frac{11}{3}}{22}=\frac{1}{6}[/tex]
[tex]\frac{b_1}{b_2}=\frac{-\frac{5}{3}}{-10}=\frac{1}{6}[/tex]
[tex]\frac{c_1}{c_2}=\frac{2}{12}=\frac{1}{6}[/tex]
[tex]\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}[/tex]
Hence, it is true.
C.[tex]10x=4y-5[/tex]
[tex]10x-4y+5=0[/tex]
[tex]\frac{\frac{11}{3}}{10}=\frac{11}{30}[/tex]
[tex]\frac{-\frac{5}{3}}{-4}=\frac{5}{12}[/tex]
[tex]\frac{11}{130}\neq\frac{5}{12}[/tex]
[tex]\frac{a_1}{a_2}\neq \frac{b_1}{b_2}[/tex]
Hence, it is not true.
D.[tex]11x=5y-6[/tex]
[tex]11x-5y+6=0[/tex]
[tex]\frac{\frac{11}{3}}{11}=\frac{1}{3}[/tex]
[tex]\frac{-\frac{5}{3}}{-5}=\frac{1}{3}[/tex]
[tex]\frac{2}{6}=\frac{1}{3}[/tex]
[tex]\therefore\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}[/tex]
Hence, it is true.
