» Substitute xx from the first equation to the second equation in terms of y.y.
\begin{gathered} \begin{cases} x = y + 5 \\ (y + 5 + 4)(y-4) = 230 \end{cases} \end{gathered}
{
x=y+5
(y+5+4)(y−4)=230
\begin{gathered} \begin{cases} x = y + 5 \\ (y + 9)(y-4) = 230 \end{cases} \end{gathered}
{
x=y+5
(y+9)(y−4)=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 - 4y + 9y - 36 = 230 \end{cases} \end{gathered}
{
x=y+5
y
2
−4y+9y−36=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y - 36 = 230 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−36=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y - 36 - 230 = 0 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−36−230=0
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y -266 = 0 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−266=0
» Solve the quadratic equation in the second equation by factoring.
y^2 + 5y - 266 = 0y
2
+5y−266=0
(y + 19)(y - 14) = 0(y+19)(y−14)=0
y + 19 = 0 \quad;\quad y - 14 = 0y+19=0;y−14=0
y = \text-19 \quad;\quad y = 14y=-19;y=14
» We can only use the positive solution since there are no negative ages. Therefore, Beth is 14 years old. Substitute it to the first equation to find Ricky's age.
\begin{gathered} \begin{cases} x = y + 5 \\ y = 14 \end{cases} \end{gathered}
{
x=y+5
y=14
\begin{gathered} \begin{cases} x = 14 + 5 \\ y = 14 \end{cases} \end{gathered}
{
x=14+5
y=14
\begin{gathered} \begin{cases} x = 19 \\ y = 14 \end{cases} \end{gathered}
{
x=19
y=14
\therefore∴ The age of Ricky is...
\large \underline{\boxed{\tt \purple{19 \: years \: old}}}
19yearsold
• While Beth is...
\large \underline{\boxed{\tt \purple{14 \: years \: old}}}
Answers & Comments
Answer:
AGES
==============================
Problem: Ricky is 5 years older than Beth. the product of his age 4 years from now and beth's age 4 years ago is 230. Find their ages.
Solution: Represent xx and yy as it is Ricky and Beth's ages respectively. Make equations on the given statements.
\begin{gathered} \begin{cases} x = y + 5 \\ (x+4)(y-4) = 230 \end{cases} \: \begin{aligned} \red{(eq. \: 1)} \\ \red{(eq. \: 2)} \end{aligned} \end{gathered}
{
x=y+5
(x+4)(y−4)=230
(eq.1)
(eq.2)
» Substitute xx from the first equation to the second equation in terms of y.y.
\begin{gathered} \begin{cases} x = y + 5 \\ (y + 5 + 4)(y-4) = 230 \end{cases} \end{gathered}
{
x=y+5
(y+5+4)(y−4)=230
\begin{gathered} \begin{cases} x = y + 5 \\ (y + 9)(y-4) = 230 \end{cases} \end{gathered}
{
x=y+5
(y+9)(y−4)=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 - 4y + 9y - 36 = 230 \end{cases} \end{gathered}
{
x=y+5
y
2
−4y+9y−36=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y - 36 = 230 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−36=230
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y - 36 - 230 = 0 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−36−230=0
\begin{gathered} \begin{cases} x = y + 5 \\ y^2 + 5y -266 = 0 \end{cases} \end{gathered}
{
x=y+5
y
2
+5y−266=0
» Solve the quadratic equation in the second equation by factoring.
y^2 + 5y - 266 = 0y
2
+5y−266=0
(y + 19)(y - 14) = 0(y+19)(y−14)=0
y + 19 = 0 \quad;\quad y - 14 = 0y+19=0;y−14=0
y = \text-19 \quad;\quad y = 14y=-19;y=14
» We can only use the positive solution since there are no negative ages. Therefore, Beth is 14 years old. Substitute it to the first equation to find Ricky's age.
\begin{gathered} \begin{cases} x = y + 5 \\ y = 14 \end{cases} \end{gathered}
{
x=y+5
y=14
\begin{gathered} \begin{cases} x = 14 + 5 \\ y = 14 \end{cases} \end{gathered}
{
x=14+5
y=14
\begin{gathered} \begin{cases} x = 19 \\ y = 14 \end{cases} \end{gathered}
{
x=19
y=14
\therefore∴ The age of Ricky is...
\large \underline{\boxed{\tt \purple{19 \: years \: old}}}
19yearsold
• While Beth is...
\large \underline{\boxed{\tt \purple{14 \: years \: old}}}
14yearsold
Answer:
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Explanation:
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