Question

A metallurgist has one alloy containing 35% titanium and another containing 64 % titanium. How many pounds of each alloy must heuse to make 52 pounds of a third alloy containing 38 % titanium? (Round to two decimal places if necessary.)Step 2 of 2: Solve the system of equations.

Answer 1

ANSWER

• 5.38 lbs of the 64% alloy

,

• 46.62 lbs of the 35% alloy

EXPLANATION

Let x be the amount of the 64% alloy and y be the amount of the 35% alloy.

We know that the two amounts add up 52 pounds,

$x+y=52$

And that the 64% of x plus the 35% of y must be 38% of the third alloy that is 52 pounds,

$0.64x+0.35y=0.38\cdot52$

Solve the first equation for y,

$y=52-x$

Replace into the second equation and solve the multiplication on the right side,

$0.64x+0.35(52-x)=19.76$

Distribute the 0.35 into the subtraction 52-x,

$\begin{gathered} 0.64x+0.35\cdot52-0.35x=19.76 \\ 0.64x+18.2-0.35x=19.76 \end{gathered}$

Add like terms,

$\begin{gathered} (0.64x-0.35x)+18.2=19.76 \\ 0.29x+18.2=19.76 \end{gathered}$

Subtract 18.2 from both sides of the equation,

$\begin{gathered} 0.29x+18.2-18.2=19.76-18.2 \\ 0.29x=1.56 \end{gathered}$

Finally divide both sides by 0.29,

$\begin{gathered} \frac{0.29x}{0.29}=\frac{1.56}{0.29} \\ x\approx5.38 \end{gathered}$

The metallurgist has to add 5.38 pounds of the 64% alloy.

To find the amount of the other alloy, we just have to replace x by 5.38 into the first equation where we solved for y before,

$y=52-5.38=46.62$

Hence, he has to add 46.62 pounds of the 35% alloy.