IS-Richmond Lesson 13: Solving Systems by Substitution

Last updated 12 months ago

8 questions

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IS-Richmond Lesson 13: Solving Systems by Substitution

Last updated 12 months ago

8 questions

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10

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10

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x+2y=8
x=−5
We can substitute x=−5 into the first equation:
−5+2y=8

-5 + 2y = 8
+5          +5
0 + 2y = 8 + 5

2y = 8 + 5
Now, solve for y
2y = 8 + 5

2y=13 (We have to eliminate the 2 in front of the y,  we do that by dividing both sides by 2)

2y = 13
2       2

y = 13/2

So, the solution to the system of equations is:
x=−5
y=13/ 2

This means the point of intersection for the two lines is  ( -5, 13/2 )

Required

10

Required

10

Required

10

Required

10

Required

10

Required

10

Required

10

x+2y=8
x=−5
We can substitute x=−5 into the first equation:
−5+2y=8

-5 + 2y = 8
+5          +5
0 + 2y = 8 + 5

2y = 8 + 5
Now, solve for y
2y = 8 + 5

2y=13 (We have to eliminate the 2 in front of the y,  we do that by dividing both sides by 2)

2y = 13
2       2

y = 13/2

So, the solution to the system of equations is:
x=−5
y=13/ 2

This means the point of intersection for the two lines is  ( -5, 13/2 )