## Precalculus (6th Edition) Blitzer

The jeweler should need $\text{96 grams}$ of $18-\text{ karat}$ gold and $\text{204 grams}$ of $12-\text{ karat}$ gold.
Let $x$ represent the grams of $18-\text{karat}$ gold, Let $y$ represent the grams of $12-\text{karat}$ gold, The amount of pure gold in each solution is found by multiplying the amount of karat by the concentration rate. This information can be organized in a table. There are two unknown quantities; therefore, a system of two independent equations relating $x$ and $y$ is set up. Amount of 75 percent gold+Amount of 28 percent gold=Amount of 25 percent gold And: Amount of pure 75 percent gold+Amount of pure 28 percent gold=Amount of pure 25 percent gold Consider the equation, \begin{align} & x+y=300 \\ & x=300-y \end{align} …… (1) And $0.75x+0.50y=174$ …… (2) Substitute $300-y$for $x$ in equation $\left( 2 \right)$ to get, \begin{align} & 0.75\left( 300-y \right)+0.50y=174 \\ & 225-0.75y+0.50y=174 \\ & -0.25y=-51 \end{align} Divide above equation by $-0.25$ to get, \begin{align} & \frac{-0.25y}{-0.25}=\frac{-51}{-0.25} \\ & y=204 \end{align} Substitute $204$for $y$ in equation $\left( 1 \right)$ to get, \begin{align} & x=300-204 \\ & x=96 \\ \end{align} Check: $\left( 96,204 \right)$ Put $x=96$and $y=204$ in the equation (1), \begin{align} \left( 96 \right)+\left( 204 \right)\overset{?}{\mathop{=}}\,300 & \\ 300=300 & \\ \end{align} And Put $x=96$and $y=204$ in the equation (2), \begin{align} 0.75\left( 96 \right)+0.50\left( 204 \right)\overset{?}{\mathop{=}}\,174 & \\ 72+102\overset{?}{\mathop{=}}\,174 & \\ 174=174 & \\ \end{align} The ordered pair $\left( 96,204 \right)$ satisfies both equations. Hence, the jeweler should need $\text{96 grams}$ of $18-\text{ karat}$ gold and $\text{204 grams}$ of $12-\text{ karat}$ gold.