A simultaneous equation, as I said before, is one that consists of two-equation made up of two variables.

To solve the word problem on a simultaneous equation, we represent each sentence with a letter. The accompanying examples illustrate the steps in solving a word problem on a simultaneous equation.

👉This post extends our posts on simultaneous equation👈

__Example 1__Four pencils and six books cost €272. Six pencils and five books cost €328. Find the co

st of one pencil and one book.

Let the cost of one pencil be $p$

Let the cost of one book be $b$

From the first sentence

$4p+6b=272$

From the second sentence

$6p+5b=328$

$4p+6b=272$.....eqn 1

$6p+5b=328$.....eqn 2

To eliminate, you will multiply eqn 1 by $6$ and eqn 2 by $4$

$6(4p+6b=272)$

$4(6p+5b=328)$

$24p+36b=1632$

$24p+20b=1312$

Substracting both equation

$24p+36b=1632$

$-(24p+20b=1312)$

$16b=320$

Divide both side by $16$

$\frac{16b}{16}=\frac{320}{16}$

$b=20$

Substituting the value of $b$ in eqn 1

$4p+6(20)=272$

$4p+120=272$

$4p=272-120$

$4p=152$

Divide both side by $4$

$\frac{4p}{4}=\frac{152}{4}$

$p=38$.

One book cost $€20$, one pen cost $€38$

__Example 2__The sum of Ada and Sam's ages is 24 years. Six years ago, Ada was three times as old as Sam. How old are they now?

Let Ada age by $x$

Let Sam age be $y$

From the first sentence

$x+y=24$.....eqn 1

From the second sentence

$(x-6)=3(y-6)$

$x-6=3y-18$

$x-3y=-18+6$

$x-3y=-12$.....eqn 2

Substracting eqn 2 from eqn 1

$x+y=24$

$-(x-3y=-12)$

4y=36

Divide both side by $3$

$\frac{4y}{4}=\frac{36}{4}$

$y=9$

Substituting $y$ in eqn 1

$x+9=24$

$x=24-9$

$x=15$

Ada is $1$ while Sam is $15$

__Example 3__The sum of the two numbers is $19$. Their difference is $5$, find the two numbers.

Let the first number be $x$

Let the second number be $y$

From the first sentence

$x+y=19$......eqn 1

From the second sentence

$x-y=5$......eqn 2

Substract eqn 2 from eqn 1

$x+y=19$

$-(x-y=5)$

$2y=14$

Divide both side by 2

$\frac{2y}{2}=\frac{14}{2}$

$y=7$

Inserting the value of $y$ in eqn 1

$x+7=19$

$x=19-7$

$x=12$

The first and second numbers are $12$ and $7$ respectively.

__Example 4__Dan's age and Josh's age add up to 25 years. Eight years ago, Dan's was twice as old as Josh. How old are they now?

Let dan age be $d$

Let Josh age be $j$

From the first sentence

$d+j=25$......eqn 1

From the second sentence

$(d-8)=2(j-8)$

$d-8=2j-16$

$d-2j=-16+8$

$d-2j=-8$....... eqn 2

Substract eqn 2 from eqn 1

$d+j=25$

$-(d-2j=-8)$

$3j=33$

$\frac{3j}{3}=\frac{33}{3}$

Divide both sides by 3

$j=11$

Substituting the value of j in eqn 1

$d+11=25$

$d=25-11$

$d=14$

Daniel is $14$ while josh's age is $11$.

__Example 5__Six books and three bags cost €234. Five books and two bags cost €184. How much does each cost?

Let the cost of the book be $x$

Let the cost of the bag be $y$

From the first sentence

$6x+3y=234$......eqn 1

From the second sentence

$5x+2y=184$.....eqn 2

For elimination sake, we multiply eqn 2 by 3 and eqn 1 by 2

$2(6x+3y=234)$

$12x+6y=468$

$3(5x+2y=184)

$15x+6y=552$

Subtracting the derived eqn

$15x+6y=552$

$-(12x+6y=468)$

$3x=84$

Divide both side by $3$

$\frac{3x}{3}=\frac{84}{3}$

$x=28$

Substituting $x$ in eqn 2

$5(28)+2y=184$

$140+2y=184$

$2y=184-140$

$2y=44$

$\frac{2y}{2}=\frac{44}{2}$

$y=22$

A book cost €28 while a bag cost €22

__Example 6__The average of two numbers is 11. Their difference is $4$. Find the numbers.

Let the first number be $x$

Let the second number be $y$

From the first sentence

$\frac{x+y}{2}=11$

$x+y=22$.....eqn 1

From the second sentence

$x-y=4$......eqn 2

Substract eqn 2 from eqn 1

$x+y=22$

$-(x-y=4)$

$2y=18$

$y=9$

Substituting the value of $y$ in eqn 2

$x-9=4$

$x=4+9$

$x=13$

The two numbers are $13$ and $9$.

__Related posts__

__Example 7__In ten years times, a father will be twice as old as his son. Ten years ago, he was six times as old as his son. How old are they now?

Let father age by $x$

Let son age be $y$

From the first sentence

$x+10=2(y+10)$

$x+10=2y+20$

$x-2y=20-10$

$x-2y=10$.....eqn 1

From the second sentence

$x-10=6(y-10)$

$x-10=6y-60$

$x-6y=-60+10$

$x-6y=-50$...... eqn 2

Subtracting eqn 2 from eqn 1

$x-2y=10$

$-(x-6y=-50)$

$4y=60$

Divide both side by $4$

$\frac{4y}{4}=\frac{60}{4}$

$y=15$

Inserting the value of $y$ in eqn 1

$x-2(15)=10$

$x-30=10$

$x=10+30$

$x=40$

The father is $40$, while his som is $15$.

Finally, we have come to the end of our series simultaneously. As usual, I will be expecting your comments in the comment box.

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