Since the 1st brand is 70% pure antifreeze
Since the 2nd brand is 95% pure antifreeze
Since we need to obtain 110 g of a mixture that contains 85% pure antifreeze
Let the quantity of the first is x and the second is y
Then
[tex]\frac{70}{100}x+\frac{95}{100}y=\frac{85}{100}(110)[/tex][tex]0.7x+0.95y=93.5\text{ (1)}[/tex][tex]x+y=110\text{ (2)}[/tex]Now let us solve the two equations to find x and y
Multiply equation (2) by -0.7
[tex]\begin{gathered} (-0.7)x+(-0.7)y=(-0.7)110 \\ -0.7x-0.7y=-77\text{ (3)} \end{gathered}[/tex]Add equations (1) and (3)
[tex]\begin{gathered} (0.7x-0.7x)+(0.95y-0.7y)=(93.5-77) \\ 0+0.25y=16.5 \\ 0.25y=16.5 \end{gathered}[/tex]Divide both sides by 0.25
[tex]\begin{gathered} \frac{0.25y}{0.25}=\frac{16.25}{0.25} \\ y=66 \end{gathered}[/tex]Substitute the value of y in equation (2) to find x
[tex]x+66=110[/tex]Subtract 66 from both sides
[tex]\begin{gathered} x+66-66=110-66 \\ x+0=44 \\ x=44 \end{gathered}[/tex]First brand: 44 gallons
Second brand: 66 gallons
