Solve the system: y = `x^2 - 6x + 5` y = 2x - 7

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The given linear equation for y is y = 2x - 7

Substituting y in the given quadratic equation `y = x^2 - 6x + 5` we get:

`2x-7= x^2 - 6x + 5`

`rArr x^2-6x+5-2x+7=0`

`rArr x^2-8x+12=0`

`rArr x^2-2x-6x+12=0`

`rArr x(x-2)-6(x-2)=0`

`rArr (x-2)(x-6)=0`

Now, using zero product property:

for (x-2)=0 we get x=2

for (x-6)=0 we get x=6

Substitute each value of x into the linear equation y = 2x - 7

So, for x=2, y=2*2-7=4-7=-3

Again, for x=6, y=2*6-7=12-7=5

Therefore, the solution set is {(2,-3),(6,5)}.

Sources:
llltkl's profile pic

Posted on

The given linear equation for y is y = 2x - 7

Substituting y in the given quadratic equation y = x^2 - 6x + 5 we get:

2x-7= x^2 - 6x + 5

rArr x^2-6x+5-2x+7=0

rArr x^2-8x+12=0

rArr x^2-2x-6x+12=0

rArr x(x-2)-6(x-2)=0

rArr (x-2)(x-6)=0

Now, using zero product property:

for (x-2)=0 we get x=2

for (x-6)=0 we get x=6

Substitute each value of x into the linear equation y = 2x - 7

So, for x=2, y=2*2-7=4-7=-3

Again, for x=6, y=2*6-7=12-7=5

Therefore, the solution set is {(2,-3),(6,5)}.

Sources:

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