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We want to find the number of solutions for the following system of equations. ${2y=4x−66x=3y+7 (I)(II) $ To determine how many solutions this system has, we will solve it by substitution. Doing so will result in one of three cases.

Result of solving by substitution | Number of solutions |
---|---|

A value for $x$ and $y$ is determined. | One solution |

An identity is found, such as $2=2.$ | Infinitely many solutions |

A contradiction is found, such as $2 =3.$ | No solution |

Thus, we need to solve the system of equations before we can make our conclusion. When solving a system of equations using substitution, there are three steps.

- Isolate a variable in one of the equations.
- Substitute the expression for that variable into the other equation and solve.
- Substitute this solution into one of the equations and solve for the value of the other variable.

$6x=3y+7$

Substitute$y=2x−3$

$6x=3(2x−3)+7$

DistrDistribute $3$

$6x=6x−9+7$

AddTermsAdd terms

$6x=6x−2$

SubEqn$LHS−6x=RHS−6x$

$0 =-2$

b

Here we have been asked to determine how many solutions the system ${x+32 y=3-6x=4y+5 (I)(II) $ has. We will first solve the system by substitution. After that we will use the following table to find the number of solutions.

Result of solving by substitution | Number of solutions |
---|---|

A value for $x$ and $y$ is determined. | One solution |

An identity is found, such as $2=2.$ | Infinitely many solutions |

A contradiction is found, such as $2 =3.$ | No solution |

$-6x=4y+5$

Substitute$x=3−32 y$

$-6(3−32 y)=4y+5$

DistrDistribute $(-6)$

$-18+4y=4y+5$

SubEqn$LHS−4y=RHS−4y$

$-18 =5$