Question Video: Finding Missing 𝑥- and 𝑦-values in Ordered Pairs Given a Linear Relation | Nagwa Question Video: Finding Missing 𝑥- and 𝑦-values in Ordered Pairs Given a Linear Relation | Nagwa

# Question Video: Finding Missing π₯- and π¦-values in Ordered Pairs Given a Linear Relation Mathematics • Second Year of Preparatory School

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In the table below some of the π₯- and π¦-values are missing from the ordered pairs that satisfy the linear relation β5π₯ + π¦ = β3. Find the missing π₯- and π¦-values.

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### Video Transcript

In the table below, some of the π₯- and π¦-values are missing from the ordered pairs that satisfy the linear relation negative five π₯ plus π¦ is equal to negative three. Find the missing π₯- and π¦-values.

We can find the missing value in each ordered pair by substituting the known π₯- or π¦-value into the equation for the linear relation. Thatβs into negative five π₯ plus π¦ is equal to negative three. We then solve for the missing π₯- or π¦-value. Alternatively, we can rearrange the equation, making the missing value the subject, and then solve.

For our first ordered pair, since we want to solve for π¦, letβs rearrange the equation so that π¦ is the subject. We can do this by adding five π₯ to both sides, which gives us π¦ is equal to negative three plus five π₯. And now substituting π₯ is equal to negative four, this gives us π¦ is equal to negative three plus five times negative four. That is, π¦ is negative three minus 20, which is negative 23. Our first ordered pair is therefore negative four, negative 23.

Now, in our second ordered pair, we know the value of π¦. That is, π¦ is equal to negative 13. So, since we want to find the value of π₯, letβs make π₯ the subject of our relation. Subtracting π¦ from both sides of negative five π₯ plus π¦ is negative three gives us negative five π₯ is equal to negative three minus π¦. And now dividing through by negative five, we have π₯ is equal to three over five plus π¦ over five. And we can rewrite this as π₯ is equal to three plus π¦ over five. And now substituting π¦ is equal to negative 13 into this equation, we have π₯ is equal to three plus negative 13 divided by five, that is, negative 10 over five, which is equal to negative two. Hence, the π₯-value in our second ordered pair is negative two. And so our ordered pair is negative two, negative 13.

Now, for our third ordered pair, we know that π₯ is equal to zero. And into our equation for π¦, this gives us π¦ is equal to negative three plus five times zero. That is, π¦ is equal to negative three. Our third ordered pair is therefore zero, negative three.

Since we again have the π₯-value β thatβs π₯ is equal to three β for our fourth ordered pair, we can use the equation for π¦ again. And so we have π¦ is equal to negative three plus five times three, that is, negative three plus 15, which is 12. Our fourth ordered pair is therefore three, 12.

And now for our fifth and final ordered pair, weβre given that π¦ has the value of 32. And substituting this into our equation for π₯, this gives us π₯ is equal to three plus 32 over five, that is, 35 divided by five, which is seven. And our final ordered pair is therefore seven, 32.

The missing values from the table are therefore π¦ is equal to negative 23, π₯ is negative two, π¦ is negative three, π¦ is 12, and π₯ is equal to seven. And hence, our ordered pairs are negative four, negative 23; negative two, negative 13; zero, negative three; three, 12; and seven, 32.

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