Question

A fruit grower can use two types of fertilizer in an orange grove, brand A and brand B.

Each bag of brand A contains $8$ pounds of nitrogen and $4$ pounds of phosphoric acid.

Each bag of brand B contains $7$ pounds of nitrogen and $7$ pounds of phosphoric acid.

Tests indicate that the grove needs $720$ pounds of nitrogen and $500$ pounds of phosphoric acid.

How many bags of each brand should be used to provide the required amounts of nitrogen and phosphoric acid?

Answer 1

Step 1: Convert the given situation to mathematical equation:

Let the required number of bags of brand A and B be $x$and $y$ respectively.

Since each bag of brand A contains $8$ pounds of nitrogen and each bag of brand B contains $7$ pounds of nitrogen:

And, the total requirement of nitrogen is $720$ pounds.

Hence according to assumption, $8x+7y=720$.

Again, the each bag of brand A contains $4$ pounds of phosphoric acid and each bag of brand B contains $7$ pounds of phosphoric acid:

And, the total requirement of phosphoric acid is $500$ pounds.

Hence according to assumption, $4x+7y=500$.

The obtain system of linear equation:

$8x+7y=720$

$4x+7y=500$

Step 2: Solve the system of equation:

Subtract the second equation from first one:

$8x+7y=720-4x-7y=-5004x=220$

Solve $4x=220$ for $x$:

$$\begin{gathered} x=\frac{220}{4} \\ =55 \end{gathered}$$

The obtain value of $x$ is $55$.

now substitute $x=55$in any of the system of equation say $8x+7y=720$ and solve of $y$:

$$\begin{gathered} 8\left(55\right)+7y=720 \\ 440+7y=720 \\ 7y=280 \\ y=40 \end{gathered}$$

The obtain value of $y$ is $40$.

Hence, the solution of the system of equation formed is $x=55$ and $y=40$.

According to assumption the number of bags required of brand A and brand B is $55$ and $40$ respectively.