PARTIAL VARIATION

Partial variation consists of two or more parts added together.

Here are some illustrations:

if $x$ is partly constant and partly varies directly as $y$, then it is written as $x=a+ky$, where ($a$ and $k$ are constant).

If $u$ is partly constant and partly varies inversely as $v$ then it is written as $u=a+\frac{k}{y}$, where $a$ and $k$ are constant.

If $b$ varies partly as $c$ and partly as the square of d, then it is written as $b=ac+kd^2$, Where $a$ and $k$ are constant.

If $x$ is partly constant and varies indirectly as the product of $y$ and $z$, then it is written as $x=a+\frac{k}{yz}$, where $a$ and $k$ are constant.

๐This post extends our posts on variation ๐

Example 1

$x$ is partly constant and partly varies as $y$. If $x=45$ when $y=10$ and $x=87$ when $y=24$. Find

1. the formula connecting $x$ and $y$

2. $x$ when $y=18$

Solution:

$x=a+ky$

$x=45$, and $y=10$

$45=a+10k$............eqn 1

$x=87$ and $y=24$

$87=a+24k$...........eqn 2

To solve simultaneously, substract eqn 1 from eqn 2

$87=a+24k$

$-(45=a+10k)$

$42=14k$

Divide both side by $14$

$\frac{42}{14}=\frac{14k}{14}$

$3=k$

$k=3$

To find $a$, we substitute $3$ for $k$ in eqn 1

$45=a+10(3)$

$45=a+30$

$45-30=a$

$15=a$

$a=15$

The formula connecting $x$ and $y$ is:

$x=15+3y$

Let's find $x$ when $y=18$

$x=15+3(18)$

$x=15+54$

$x=69$

Example 2

A variable ($x$) varies directly as $y$ and inversely as the square of $y$. Given that $x=11$ when $y=2$ and $x=\frac{629}{25}$ when $y=5$, find $x$ when $y=7$

Solution:

$x=ay+\frac{k}{y^2}$

When $x=11$, $y=2$

$11=2a+\frac{k}{2^2}$

$11=2a+\frac{k}{4}$

Taking the L.C.M

$11=\frac{8a+k}{4}$

Cross multiply

$44=8a+k$.......eqn 1

When $x=\frac{629}{25}$, $y=5$

$\frac{629}{25}=5a+\frac{k}{5^2}$

$\frac{629}{25}=5a+\frac{k}{25}$

Taking the L.C.M

$\frac{629}{25}=\frac{125a+k}{25}$

For simplicity, let's Multiply both side by $25$

$\frac{629}{25}\times25=\frac{125a+k}{25}\times 25$

$629=125a+k$.....eqn 2

Subtracting eqn 1 from eqn 2

$629=125a+k$

$-(44=8a+k$

$585=117a$

Divide both sides by $117$

$\frac{585}{117}=\frac{117a}{117}$

$a=5$

Substituting $5$ for a in eqn 1

$44=8(5)+k$44=40+k44-40=k4=kk=4x=5y+\frac{4}{y^2}$Now, let's find$x$when$y=7x=5(7)+\frac{4}{7^2}x=35+\frac{4}{49}$Taking the L.C.M$x=\frac{1715+4}{49}x=\frac{1719}{49}x=35.1$Example 3$b$is partly constant and partly varies directly as$c$when$c=5$,$b=7$, and when$c=7$,$b=7$1. Find the equation connecting$x$and$y$2. Find$b$when$c=11$Solution:$b=a+kcc=5$,$b=77=a+k5$.......eqn 1$c=7$,$b=78=a+k7$.......eqn 2 Substract eqn 1 from eqn 2$8=a+k7-(7=a+k5)1=2k$Divide both side by$2\frac{1}{2}=\frac{2k}{2}\frac{1}{2}=kk=\frac{1}{2}$To solve for$a$, let's substitute the value of$k$into eqn 1$7=a+(\frac{1}{2})57=a+\frac{5}{2}$Taking the L.C.M$7=\frac{2a+5}{2}$Cross multiply$7\times2=2a+514=2a+514-5=2a9=2a$Divide both side by$2\frac{9}{2}=\frac{2a}{2}\frac{9}{2}=aa=\frac{9}{2}$Therefore, the equation connecting$b$and$c$is:$b=\frac{9}{2}+\frac{c}{2}$Finding$b$when$c=11b=\frac{9}{2}+\frac{11}{2}$Taking the L.C.M$b=\frac{9+11}{2}b=\frac{20}{2}b=10$Related posts Example 4$p$is partly constant and partly varies indirectly as the cube root of$q$. When$q=125$,$p=35$and when$q=8000$,$p=20$. Find 1. The relation between$p$and$q$2. Find$q$when$p=215$Solution:$p=a+\frac{k}{\sqrt[3]{q}}p=35$,$q=125$,$35=a+\frac{k}{\sqrt[3]{125}}35=a+\frac{k}{5}$Taking the L.C.M$35=\frac{5a+k}{5}$Cross multiply$35\times5=5a+k175=5a+k$...........eqn 1$q=8000$,$p=202t0=a+\frac{k}{\sqrt[3]{8000}}20=a+\frac{k}{20}$Taking the L.C.M of both sides$20=\frac{20a+k}{20}$Cross multiply$20\times20=20a+k400=20a+k$......... eqn 2 Substract eqn 1 from eqn 2$400=20a+k-(175=5a+k225=15a$Divide both side by 15$\frac{225}{15}=\frac{15a}{15}15=aa=15$To solve for$k$, you substitute$15$for$k$in eqn 1$175=5(15)+k175=75+k175-75=k100=kk=100$The relation between$p$and$q$is$p=15+\frac{100}{\sqrt[3]{q}}$Solving for$q$when$p=215215=15+\frac{100}{\sqrt[3]{q}}$Taking the L.C.M$215=\frac{15\sqrt[3]{q}+100}{\sqrt[3]{q}}$Cross multiply$215\times\sqrt[3]{q}=15\sqrt[3]{q}+100215\sqrt[3]{q}=15\sqrt[3]{q}+100$Collect like terms$215\sqrt[3]{q}_15\sqrt{3}{q}=100200\sqrt[3]{q}=100$Divide both side by$200\frac{200\sqrt[3]{q}}{200}=\frac{100}{200}\sqrt[3]{q}=\frac{1}{2}$To get rid of cube root, we add cube both sides.$(\sqrt[3]{q})^3=(\frac{1}{2})^3q=\frac{1}{8}\$

There you have it. We would continue our series on variations when we explore Word problems on direct and indirect variations.

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1 comment

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