Use substitution to determine whether the given ordered pairs are solutions of the given equation. $$ (-3.5,-1.5),(-3,1) ; x^{2}+y^{2}=10 $$ Is $$ (-3.5,-1.5) $$ a solution of the given equation? No Yes Is $$ (-3,1) $$ a solution of the given equation? Yes No

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Substitute $$ x=-3.5,y=-1.5 $$ into the expression $$ x^{2}+y^{2}=10 $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$x^{2}+y^{2}=10$$
- step1: Substitute:
  $$\left(-3.5ight)^{2}+\left(-1.5ight)^{2}=10$$
- step2: Convert the expressions:
  $$\left(-\frac{7}{2}ight)^{2}+\left(-1.5ight)^{2}=10$$
- step3: Convert the expressions:
  $$\left(-\frac{7}{2}ight)^{2}+\left(-\frac{3}{2}ight)^{2}=10$$
- step4: Add the numbers:
  $$\frac{29}{2}=10$$
- step5: Multiply both sides of the equation by LCD:
  $$\frac{29}{2}	imes 2=10	imes 2$$
- step6: Simplify the equation:
  $$29=20$$
- step7: Check the equality:
  $$	extrm{false}$$
Substitute $$ x=-3,y=1 $$ into the expression $$ x^{2}+y^{2}=10 $$.
Evaluate the expression by following steps:
- step0: Evaluate:
  $$x^{2}+y^{2}=10$$
- step1: Substitute:
  $$\left(-3ight)^{2}+1^{2}=10$$
- step2: Evaluate the power:
  $$\left(-3ight)^{2}+1=10$$
- step3: Add the numbers:
  $$10=10$$
- step4: Check the equality:
  $$	extrm{true}$$
The ordered pair $$ (-3.5,-1.5) $$ is not a solution of the equation $$ x^{2}+y^{2}=10 $$.

The ordered pair $$ (-3,1) $$ is a solution of the equation $$ x^{2}+y^{2}=10 $$.
$$ (-3.5, -1.5) $$ is not a solution of $$ x^{2} + y^{2} = 10 $$. $$ (-3, 1) $$ is a solution of $$ x^{2} + y^{2} = 10 $$.

Use substitution to determine whether the given ordered pairs are solutions of the given equation. \( (-3.5,-1.5),(-3,1) ; x^{2}+y^{2}=10 \) Is \( (-3.5,-1.5) \) a solution of the given equation? No Yes Is \( (-3,1) \) a solution of the given equation? Yes No

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