Question

example 3 apply the substitution method
chemistry ms. washington is preparing a hydrochloric acid (hcl) solution. she will need 300 milliliters of a 5% hcl solution for her class to use during a lab. if she has a 3.5% hcl solution and a 7% hcl solution, how much of each solution should she use in order to make the solution needed?
step 1 write two equations in two variables.
let x be the amount of 3.5% solution and y be the amount of 7% solution.
$x + y = 300$ equation 1
$0.035x + 0.07y = 0.05(300)$ equation 2
step 2 solve one equation for one of the variables.
$x + y = 300$ equation 1
$x = __ + 300$ subtract y from each side.
step 3 substitute the resulting expression and solve.
$0.035x + 0.07y = 15$ equation 2
$0.035 ___ + 0.07y = 15$ $x = -y + 300$
$__y + __ + 0.07y = 15$ distributive property
$0.035y = 4.5$ simplify.
$y approx ___$ divide each side by 0.035.
go online you can complete an extra example online.
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Explanation:

Step1: Resolver la ecuación 1 para \(x\)

Dado \(x + y=300\), restando \(y\) de ambos lados, obtenemos \(x=-y + 300\)

Step2: Sustituir \(x=-y + 300\) en la ecuación 2

La ecuación 2 es \(0.035x+0.07y = 15\). Sustituyendo \(x\) tenemos:
\(0.035(-y + 300)+0.07y=15\)
Aplicando la propiedad distributiva \(a(b + c)=ab+ac\) (aquí \(a = 0.035\), \(b=-y\), \(c = 300\)):
\(-0.035y+10.5 + 0.07y=15\)

Step3: Simplificar la ecuación resultante

Combinando términos semejantes (\(-0.035y+0.07y=(0.07 - 0.035)y=0.035y\)):
\(0.035y+10.5 = 15\)
Restando \(10.5\) de ambos lados: \(0.035y=15 - 10.5=4.5\)

Step4: Resolver para \(y\)

Dividiendo ambos lados de la ecuación \(0.035y = 4.5\) por \(0.035\):
\(y=\frac{4.5}{0.035}=\frac{4500}{35}=\frac{900}{7}\approx128.57\)

Step5: Resolver para \(x\)

Sustituyendo \(y=\frac{900}{7}\) en la ecuación \(x=-y + 300\)
\(x=-\frac{900}{7}+300=\frac{-900 + 2100}{7}=\frac{1200}{7}\approx171.43\)

Answer:

Ella debe usar aproximadamente \(x=\frac{1200}{7}\approx171.43\) mililitros de la solución \(3.5\%\) y \(y = \frac{900}{7}\approx128.57\) mililitros de la solución \(7\%\)