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1 year ago

What is the value of x in the equation 1/5x-2/3y = 30, when y = 15?

Mathematics

2 answers:

Nataly [62]1 year ago

8 0

Answer:

x = 200

Step-by-step explanation:

Given equation:

$\sf \dfrac{1}{5}x-\dfrac{2}{3}y=30$

Steps:

1. Substitute 15 as the value of y in the equation:

$\sf \dfrac{1}{5}x-\dfrac{2}{3}(15)=30\ \textsf{[ multiply ]}\\\\\Rightarrow \dfrac{1}{5}x-\dfrac{30}{3}=30\ \textsf{[ simplify ]}\\\\\Rightarrow \dfrac{1}{5}x-10=30$

2. Add 10 to both sides:

$\sf \dfrac{1}{5}x-10+10=30+10\\\\\Rightarrow \dfrac{1}{5}x=40$

3. Multiply both sides by 5 to isolate x:

$\sf 5\left(\dfrac{1}{5}\right)x=5(40)\\\\\Rightarrow x=200$

4. Check your work:

$\sf \dfrac{1}{5}x-\dfrac{2}{3}y=30\ \textsf{[ substitute 200 for x, and 15 for y ]}\\\\\dfrac{1}{5}(200)-\dfrac{2}{3}(15)=30\ \textsf{[ multiply ]}\\\\\dfrac{200}{5}-\dfrac{30}{3}=30\ \textsf{[ divide ]}\\\\40-10=30\ \textsf{[ subtract ]}\\\\30=30\ \checkmark$

Learn more here:

brainly.com/question/12965239

Alika [10]1 year ago

3 0

Hey there!

Answer :

$\boxed{\bold{\green{ x = 200}}}$

$\\$

Explanation:

Given expression :

$\sf{ \frac{1}{5} \green{x} - \frac{2}{3} \orange{y} = 30 }$

1) Substitue 15 for $\sf{\orange{y}}$ :

$\sf{ \frac{1}{5} \green{x} - \frac{2}{3} \orange{ \times 15} = 30 } \\ \\ \implies\sf{ \frac{1}{5} \green{x} - \frac{30}{3} = 30 } \\ \\ \implies\sf{ \frac{1}{5} \green{x} - 10= 30 }$

$\\$

2) Solve for x :

$\sf{ \frac{1}{5} \green{x} - 10= 30 }$

⇢Add 10 to both sides of the equation :

$\sf{ \frac{1}{5} \green{x} - 10 \: \bold{+ \: 10}= 30 \: \bold{+ \: 10 }} \\ \\ \implies \sf{\frac{1}{5} \green{x} \: = 40 } \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

⇢Multiply both sides of the equation by 5:

$\Big( \dfrac{1}{5} \green{x}\Big) \: \bold{ \times \: 5}=40 \: \bold{ \times \: 5} \\ \\ \implies \sf{\dfrac{5}{5} \green{x} \: = 200} \\ \\ \implies \green{ \boxed{\sf{x = 200}}}$

$\\$

3) Let's check our answer by replacing $\green{x}$ with 200 and $\orange{y}$ with 15 :

$\sf{\dfrac{1}{5}\overbrace{\green{\times \: 200}}^{\green{x}} - \dfrac{2}{3}\underbrace{\orange{\times \:15}}_{\orange{y}}= \dfrac{200}{5} - \dfrac{30}{30} = 40 - 10 = \boxed{ 30} }$