Algebra

Ontario 09 Applied (MFM1P)

Substitution resulting in an equation

Lesson

In Using Your Formulae, we looked at how to substitute values into common equations to evaluate the subject of the formulae. Sometimes we don't know all of the values, and we just want to substitute the ones that we do know. This leaves us with a simpler equation that involves less variables.

In Physics, Ohm's Law states that $V=I\times R$`V`=`I`×`R` for a conductor, where $V$`V` is the voltage (in Volts), $I$`I` is the current (in Amps) and $R$`R` is the resistance (in Ohms). If a particular object in a circuit has a resistance of $100$100 Ohms, write the voltage in terms of the current.

Think: We want to substitute the given value of $R$`R`, so that we are left with an equation using only $V$`V` and $I$`I`.

Do: We have that $R=100$`R`=100, so $V=I\times100$`V`=`I`×100, or $V=100I$`V`=100`I`.

The area $A$`A` of a trapezoid is given by the equation $A=\frac{1}{2}h\left(a+b\right)$`A`=12`h`(`a`+`b`), where $a$`a` and $b$`b` are the lengths of the parallel sides and $h$`h` is the perpendicular distance between them. If the perpendicular height of a trapezoid is $12$12 cm, write an equation for the area in terms of the lengths $a$`a` and $b$`b`.

Think: We want to substitute the given value of $h$`h`, so that we are left with an equation for $A$`A` using only $a$`a` and $b$`b`. We will also be able to simplify this time.

Do: We have that $h=12$`h`=12, so $A=\frac{1}{2}\times12\left(a+b\right)$`A`=12×12(`a`+`b`), which simplifies to $A=6\left(a+b\right)$`A`=6(`a`+`b`).

The area of a rectangle, $A$`A`, is given by the formula $A=L\times W$`A`=`L`×`W`, where $L$`L` is the length and $W$`W` is the width.

If the length of a rectangle is $15$15 cm, select the expression that gives the area of the rectangle in terms of the width.

$A=\frac{15}{W}$

`A`=15`W`A$A=\frac{15}{L}$

`A`=15`L`B$A=15L$

`A`=15`L`C$A=15W$

`A`=15`W`D$W=15A$

`W`=15`A`E$A=15+W$

`A`=15+`W`F$A=\frac{15}{W}$

`A`=15`W`A$A=\frac{15}{L}$

`A`=15`L`B$A=15L$

`A`=15`L`C$A=15W$

`A`=15`W`D$W=15A$

`W`=15`A`E$A=15+W$

`A`=15+`W`F

The perimeter, $P$`P`, of a triangle with sides of lengths $x$`x`, $y$`y` and $z$`z` is given by the formula $P=x+y+z$`P`=`x`+`y`+`z`.

A triangle has two known side lengths, $x=5$`x`=5 cm and $y=8$`y`=8 cm. Express the perimeter of the triangle in terms of the unknown side length $z$`z`.

The equation of a straight line is given by the formula $y=mx+c$`y`=`m``x`+`c`, where $m$`m` is the slope and $c$`c` is the value at the $y$`y`-intercept.

If the slope of a straight line is $8$8, select the expression that gives the equation of the straight line in terms of $x$`x` and $c$`c`.

$y=x+c$

`y`=`x`+`c`A$xc=8y$

`x``c`=8`y`B$c=8x$

`c`=8`x`C$y=8xc$

`y`=8`x``c`D$y=8x$

`y`=8`x`E$y=8x+c$

`y`=8`x`+`c`F$y=x+c$

`y`=`x`+`c`A$xc=8y$

`x``c`=8`y`B$c=8x$

`c`=8`x`C$y=8xc$

`y`=8`x``c`D$y=8x$

`y`=8`x`E$y=8x+c$

`y`=8`x`+`c`F

Substitute into algebraic equations and solve for one variable in the first degree