b must be equal to -6 for infinitely many solutions for system of equations [tex]y = 6x + b[/tex] and [tex]-3 x+\frac{1}{2} y=-3[/tex]
Solution:
Need to calculate value of b so that given system of equations have an infinite number of solutions
[tex]\begin{array}{l}{y=6 x+b} \\\\ {-3 x+\frac{1}{2} y=-3}\end{array}[/tex]
Let us bring the equations in same form for sake of simplicity in comparison
[tex]\begin{array}{l}{y=6 x+b} \\\\ {\Rightarrow-6 x+y-b=0 \Rightarrow (1)} \\\\ {\Rightarrow-3 x+\frac{1}{2} y=-3} \\\\ {\Rightarrow -6 x+y=-6} \\\\ {\Rightarrow -6 x+y+6=0 \Rightarrow(2)}\end{array}[/tex]
Now we have two equations
[tex]\begin{array}{l}{-6 x+y-b=0\Rightarrow(1)} \\\\ {-6 x+y+6=0\Rightarrow(2)}\end{array}[/tex]
Let us first see what is requirement for system of equations have an infinite number of solutions
If [tex]a_{1} x+b_{1} y+c_{1}=0[/tex] and [tex]a_{2} x+b_{2} y+c_{2}=0[/tex] are two equation
[tex]\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/tex] then the given system of equation has no infinitely many solutions.
In our case,
[tex]\begin{array}{l}{a_{1}=-6, \mathrm{b}_{1}=1 \text { and } c_{1}=-\mathrm{b}} \\\\ {a_{2}=-6, \mathrm{b}_{2}=1 \text { and } c_{2}=6} \\\\ {\frac{a_{1}}{a_{2}}=\frac{-6}{-6}=1} \\\\ {\frac{b_{1}}{b_{2}}=\frac{1}{1}=1} \\\\ {\frac{c_{1}}{c_{2}}=\frac{-b}{6}}\end{array}[/tex]
As for infinitely many solutions [tex]\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}[/tex]
[tex]\begin{array}{l}{\Rightarrow 1=1=\frac{-b}{6}} \\\\ {\Rightarrow6=-b} \\\\ {\Rightarrow b=-6}\end{array}[/tex]
Hence b must be equal to -6 for infinitely many solutions for system of equations [tex]y = 6x + b[/tex] and [tex]-3 x+\frac{1}{2} y=-3[/tex]
